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Navier-Stokes, stationary, two dimensions, non-symmetric, laminar wake
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\begin{document}

\title{Leading order down-stream asymptotics of non-symmetric stationary
Navier-Stokes flows in two dimensions}
\author{Fr\'{e}d\'{e}ric Haldi\\{\small D\'{e}partement de Physique Th\'{e}orique}\\{\small Universit\'{e} de Gen\`{e}ve, Switzerland}
\and Peter Wittwer\thanks{Supported in part by the Fonds National Suisse.}\\{\small D\'{e}partement de Physique Th\'{e}orique}\\{\small Universit\'{e} de Gen\`{e}ve, Switzerland}\\{\small peter.wittwer@physics.unige.ch}}
\maketitle

\begin{abstract}
We consider stationary solutions of the incompressible{\normalsize \ }%
Navier-Stokes equations in two dimensions. We give a detailed description of
the fluid flow in a half-plane through the construction of an inertial
manifold for the dynamical system that one obtains when using the coordinate
along the flow as a time.

\end{abstract}

\section{Introduction}

We consider, in $d=2$ dimensions, the time independent incompressible
Navier-Stokes equations
\begin{align}
-(\mathbf{u}\cdot\nabla)\mathbf{u}+\Delta\mathbf{u}-\mathbf{\nabla}p &
=0~,\label{navier}\\
\nabla\cdot\mathbf{u} &  =0~,\label{div0}%
\end{align}
in the half-space $\Omega=\{(x,y)\in\mathbf{R}^{2}\left\vert {}\right.
x>1\}$. We are interested in modeling the situation where fluid enters
$\Omega$ through the surface $\Sigma=\{(x,y)\in\mathbf{R}^{2}\left\vert
{}\right.  x=1\}$ and where the fluid flows at infinity parallel to the
$x$-axis at a nonzero constant speed $\mathbf{u}_{\mathbf{\infty}}\equiv
(1,0)$. We therefore impose the boundary conditions
\begin{align}
\lim_{\substack{x\geq1 \\x^{2}+y^{2}\rightarrow\infty}}\mathbf{u}(\mathbf{x})
& =\mathbf{u}_{\mathbf{\infty}}~,\label{bc001}\\
\left.  \mathbf{u}\right\vert _{\Sigma}  & =\mathbf{u}_{\mathbf{\infty}%
}+\mathbf{u}_{\ast}~,\label{bc002}%
\end{align}
with $\mathbf{u}_{\ast}=(u_{\ast},v_{\ast})$ in a certain set of vector fields
$\mathcal{S}$ to be defined later on, and satisfying $\lim_{\left\vert
y\right\vert \rightarrow\infty}\mathbf{u}_{\ast}(y)=0$.

The above problem has been studied in detail in \cite{pw001}, \cite{pw002},
for the special case where $u_{\ast}(y)=u_{\ast}(-y)$ and $v_{\ast
}(y)=-v_{\ast}(-y)$. As a consequence the discussion could there be restricted
to the case of symmetric vector fields $\mathbf{u=u}_{\infty}+(u,v)$,
\textit{i.e.}, to functions $u$, $v$ and $p$ satisfying $u(x,y)=u(x,-y)$,
$v(x,y)=-v(x,-y)$ and $p(x,y)=p(x,-y)$ for all $x\geq1$, and this symmetry
property was extensively used in the proofs. In order to get rid of this
limitation one is forced to study the nonlinearity in (\ref{navier}) in much
more detail than in \cite{pw001}, \cite{pw002}. This makes the estimates
somewhat lengthy, since many different terms have to be analyzed, but also
simpler, since less information needs to be encoded in the function spaces.

The following theorem is our main result.

\begin{theorem}
\label{theorem0}Let $\Sigma$ and $\Omega$ be as defined above. Then, for each
$\mathbf{u}_{\ast}=(u_{\ast},v_{\ast})$ in a certain set of vector fields
$\mathcal{S}$ to be defined later on, there exist a vector field
$\mathbf{u}=\mathbf{u}_{\infty}+(u,v)$ and a function $p$ satisfying the
Navier-Stokes equations (1) and (2) in $\Omega$ subject to the boundary
conditions (3) and (4). Furthermore,
\begin{align}
\lim\limits_{x\rightarrow\infty}x^{1/2}\left(  \sup\limits_{y\in\mathbf{R}%
}\left\vert \left(  u-u_{as}\right)  (x,y)\right\vert \right)   &
=0~,\label{th2}\\
\lim\limits_{x\rightarrow\infty}x\left(  \sup\limits_{y\in\mathbf{R}%
}\left\vert \left(  v-v_{as}\right)  (x,y)\right\vert \right)   &
=0~,\label{th3}%
\end{align}
where%
\begin{align}
u_{as}(x,y) &  =\frac{c}{2\sqrt{\pi}}\frac{1}{\sqrt{x}}e^{-\frac{y^{2}}{4x}%
}+\frac{d}{\pi}\frac{x}{x^{2}+y^{2}}+\frac{b}{\pi}\frac{y}{x^{2}+y^{2}%
}~,\label{xxx}\\
v_{as}(x,y) &  =\frac{c}{4\sqrt{\pi}}\frac{y}{x^{3/2}}e^{-\frac{y^{2}}{4x}%
}+\frac{d}{\pi}\frac{y}{x^{2}+y^{2}}-\frac{b}{\pi}\frac{x}{x^{2}+y^{2}%
}~,\label{yyy}%
\end{align}
with
\begin{align}
b &  =\lim\limits_{k\rightarrow0^{+}}\int_{\mathbf{R}}\sin(ky)~u_{\ast
}(y)~dy~,\label{bbb}\\
d &  =\lim\limits_{k\rightarrow0^{+}}\int_{\mathbf{R}}\sin(ky)~v_{\ast
}(y)~dy~,\label{ddd}\\
c &  =\lim\limits_{k\rightarrow0}\int_{\mathbf{R}}\cos(ky)~u_{\ast
}(y)~dy~-d~.\label{ccc}%
\end{align}

\end{theorem}

\bigskip

A proof of this theorem is given in Section 8.

\begin{remark}
The integrals in (\ref{bbb}), (\ref{ddd}) and (\ref{ccc}) have to be
understood in the $(C,\delta)$-sense, with $0<\delta\leq1$ (see e.g.
\cite{Titchmarsh}, Theorem 15). Namely, let $C(\delta,R)=$ $(1-\left\vert
y\right\vert /R)^{\delta}$. Then the exact version of (\ref{bbb}) is
\[
b=\lim\limits_{k\rightarrow0_{+}}\left(  \lim\limits_{R\rightarrow\infty}%
\int_{-R}^{R}C(\delta,R)\sin(ky)u_{\ast}(y)~dy\right)  ~,
\]
and accordingly for the other cases.
\end{remark}

The set $\mathcal{S}$\ in Theorem \ref{theorem0} will be specified in Section
8, once appropriate function spaces have been introduced.\textit{\ }For an
interpretation of the results see \cite{pw001}, \cite{pw002}. For related
results see \cite{gald98a}, \cite{gald98b} and \cite{vanbaalen001}. For an
application of the above results for an efficient numerical implementation of
two-dimensional stationary exterior flow problems see \cite{vincent001}.

The rest of this paper is organized as follows. In Section 2 and Section 3 we
rewrite equation (\ref{navier}) and (\ref{div0}) as a dynamical system with
the coordinate parallel to the flow playing the role of time. The discussion
will be formal. At the end of the discussion we get a set of integral
equations. In Sections 4 and 5 we then prove that these integral equations
admit a solution. This solution is analyzed in some detail in Section 6 and
Section 7. In Section 8 we finally prove Theorem \ref{theorem0} by using the
results from Sections 4-7.

\section{The dynamical system}

Let $\mathbf{u}=\mathbf{u}_{\infty}+(u,v)$. Then, the equations (\ref{navier}%
), (\ref{div0}) are equivalent to
\begin{align}
\omega & =\partial_{x}v-\partial_{y}u~,\nonumber\\
0  & =-(\mathbf{u\cdot\nabla)}\omega\mathbf{+\Delta}\omega~,\nonumber\\
0  & =\partial_{x}u+\partial_{y}v~,\label{ww}%
\end{align}
since the pressure $p$ is uniquely determined (modulo an additive constant) by
solving%
\begin{equation}
\Delta p=\mathbf{\nabla\cdot}\left(  (\mathbf{u\cdot\nabla})\mathbf{u}%
-\mathbf{\Delta u}\right) \label{lp}%
\end{equation}
in $\Omega$ with the boundary condition
\begin{equation}
\partial_{x}p=-(\mathbf{u}\cdot\nabla)u+\Delta u\label{bp}%
\end{equation}
on $\Sigma$, once that (\ref{ww}) has been solved. The function $\omega$ is
the vorticity of the fluid.

Provided $\partial_{x}u+\partial_{y}v=0$, we have that
\begin{equation}
u\partial_{x}\omega+v\partial_{y}\omega=\partial_{x}(u\omega)+\partial
_{y}(v\omega)\equiv q~,\label{nnn}%
\end{equation}
and - for reasons which are not obvious to us - it turns out to be important
to discuss the nonlinearity $q$ as represented by the expression on the
\textit{r.h.s.} of (\ref{nnn})\footnote{In \cite{haldi001} part of the results
of this paper were proved by using the expression on the \textit{l.h.s. }of
(\ref{nnn}). That approach turned out to be much more complicated than the
present one.}.

The main idea underlying the tools developed in this paper is to consider the
coordinate parallel to the flow as a time coordinate \cite{batchelor}. Let
$\eta=\partial_{x}\omega$. Then, the equations (\ref{ww}) are equivalent to%
\begin{align}
\partial_{x}\omega & =\eta~,\nonumber\\
\partial_{x}\eta & =\eta-\partial_{y}^{2}w+q~,\nonumber\\
\partial_{x}u  & =-\partial_{y}v~,\nonumber\\
\partial_{x}v  & =\partial_{y}u+\omega~.\label{ds}%
\end{align}
Let
\[
\omega(x,y)=\frac{1}{2\pi}\int_{\mathbf{R}}dk~e^{-iky}\hat{\omega}(k,x)~,
\]
and accordingly for the other functions. For (\ref{ds}) we then get (for
simplicity we drop the hats and use in Fourier space $t$ instead of $x$ for
the \textquotedblleft time\textquotedblright-variable) the dynamical system
\begin{align}
\dot{\omega}  & =\eta~,\nonumber\\
\dot{\eta}  & =\eta+k^{2}\omega+q~,\nonumber\\
\dot{u}  & =ikv~,\nonumber\\
\dot{v}  & =-iku+\omega~,\label{dd}%
\end{align}
where
\begin{equation}
q=\partial_{t}q_{0}-ikq_{1}~,\label{qq}%
\end{equation}
with%
\begin{align}
q_{0}  & =\frac{1}{2\pi}\left(  u\ast\omega\right)  ~,\label{q0o}\\
q_{1}  & =\frac{1}{2\pi}\left(  v\ast\omega\right)  ~.\label{q1o}%
\end{align}
The equations (\ref{dd}) are of the form $\mathbf{\dot{z}}=L\mathbf{z}%
+\mathbf{q}$, with $\mathbf{z}=(\omega,\eta,u,v)$, $\mathbf{q}=(0,q,0,0)$ and
\[
L(k)=\left(
\begin{array}
[c]{cccc}%
0 & 1 & 0 & 0\\
k^{2} & 1 & 0 & 0\\
0 & 0 & 0 & ik\\
1 & 0 & -ik & 0
\end{array}
\right)  ~.
\]
The matrix $L(k)$ can be diagonalized. Namely, let $\sigma(k)\equiv
\mathrm{signum}(k)$, and define $\Lambda_{0},$ $\Lambda_{+}$ and $\Lambda_{-}$
by
\begin{align*}
\Lambda_{0}(k)  & =\sqrt{1+4k^{2}}~,\\
\Lambda_{+}(k)  & =\frac{1+\Lambda_{0}(k)}{2}~,\\
\Lambda_{-}(k)  & =\frac{1-\Lambda_{0}(k)}{2}~.
\end{align*}
Let $\mathbf{z=}S\mathbf{\zeta}$ with
\[
S(k)=\left(
\begin{array}
[c]{cccc}%
1 & 1 & 0 & 0\\
\Lambda_{+} & \Lambda_{-} & 0 & 0\\
-\frac{i}{k}\Lambda_{-} & -\frac{i}{k}\Lambda_{+} & 1 & 1\\
1 & 1 & -i\sigma & i\sigma
\end{array}
\right)  ~.
\]
Then $\mathbf{\dot{\zeta}}=D\mathbf{\zeta}+S^{-1}\mathbf{q}$ with
\[
S^{-1}(k)=\left(
\begin{array}
[c]{cccc}%
\begin{array}
[c]{c}%
-\frac{\Lambda_{-}}{\Lambda_{0}}\\
~
\end{array}
&
\begin{array}
[c]{c}%
\frac{1}{\Lambda_{0}}\\
~
\end{array}
&
\begin{array}
[c]{c}%
0\\
~
\end{array}
&
\begin{array}
[c]{c}%
0\\
~
\end{array}
\\%
\begin{array}
[c]{c}%
\frac{\Lambda_{+}}{\Lambda_{0}}\\
~
\end{array}
&
\begin{array}
[c]{c}%
-\frac{1}{\Lambda_{0}}\\
~
\end{array}
&
\begin{array}
[c]{c}%
0\\
~
\end{array}
&
\begin{array}
[c]{c}%
0\\
~
\end{array}
\\%
\begin{array}
[c]{c}%
-\frac{i}{2}(\sigma-\frac{1}{k})\\
~
\end{array}
&
\begin{array}
[c]{c}%
-\frac{1}{2}\frac{i}{k}\\
~
\end{array}
&
\begin{array}
[c]{c}%
\frac{1}{2}\\
~
\end{array}
&
\begin{array}
[c]{c}%
\frac{1}{2}i\sigma\\
~
\end{array}
\\
~~\frac{i}{2}(\sigma+\frac{1}{k}) & -\frac{1}{2}\frac{i}{k} & \frac{1}{2} &
-\frac{1}{2}i\sigma
\end{array}
\right)  ~,
\]
and $D=S^{-1}LS$ a diagonal matrix with diagonal entries $\Lambda_{+}$,
$\Lambda_{-}$, $\left\vert k\right\vert $, and $-\left\vert k\right\vert $.
Note that $\Lambda_{+}(k)\geq1$ and $\Lambda_{-}(k)\leq0$ and $\Lambda
_{-}(k)\approx-k^{2}$ for small values of $k.$ Let $\mathbf{\zeta}$
$=(\omega_{+},\omega_{-},u_{+},u_{-}).$ Using the definitions we find that
(\ref{dd}) is equivalent to
\begin{align}
\dot{\omega}_{+}  & =\Lambda_{+}\omega_{+}+\frac{1}{\Lambda_{0}}q~,\nonumber\\
\dot{\omega}_{-}  & =\Lambda_{-}\omega_{-}-\frac{1}{\Lambda_{0}}q~,\nonumber\\
\dot{u}_{+}  & =\left\vert k\right\vert u_{+}-\frac{1}{2}\frac{i}%
{k}q~,\nonumber\\
\dot{u}_{-}  & =-\left\vert k\right\vert u_{-}-\frac{1}{2}\frac{i}%
{k}q~,\label{ddpm}%
\end{align}
with $q$ as defined in (\ref{qq}). For convenience later on we also write
$\mathbf{z=}S\mathbf{\zeta}$ in component form. Namely,
\begin{align}
\omega & =\omega_{+}+\omega_{-}~,\nonumber\\
\eta & =\Lambda_{+}\omega_{+}+\Lambda_{-}\omega_{-}~,\nonumber\\
u  & =-\frac{i}{k}\Lambda_{-}\omega_{+}-\frac{i}{k}\Lambda_{+}\omega_{-}%
+u_{+}+u_{-}~,\nonumber\\
v  & =\omega_{+}+\omega_{-}-i\sigma u_{+}+i\sigma u_{-}~.\label{uv}%
\end{align}


\section{The integral equations}

To solve (\ref{ddpm}) we convert it into an integral equation. The $+$-modes
are unstable (remember that $\Lambda_{+}(k)\geq1$) and we therefore have to
integrate these modes backwards in time starting with $\omega_{+}%
(k,\infty)\equiv u_{+}(k,\infty)\equiv0$ (see \cite{Gallay}). We get
\begin{align}
\omega_{+}(k,t)  & =-\frac{1}{\Lambda_{0}}\int_{t}^{\infty}e^{\Lambda
_{+}(t-s)}q(k,s)~ds~,\label{pm1}\\
\omega_{-}(k,t)  & =\tilde{\omega}_{-}^{\ast}(k)e^{\Lambda_{-}(t-1)}-\frac
{1}{\Lambda_{0}}\int_{1}^{t}e^{\Lambda_{-}(t-s)}q(k,s)~ds~,\label{pm2}\\
u_{+}(k,t)  & =\frac{1}{2}\frac{i}{k}\int_{t}^{\infty}e^{\left\vert
k\right\vert (t-s)}q(k,s)~ds~,\label{pm3}\\
u_{-}(k,t)  & =\tilde{u}_{-}^{\ast}(k)e^{-\left\vert k\right\vert (t-1)}%
-\frac{1}{2}\frac{i}{k}\int_{1}^{t}e^{-\left\vert k\right\vert (t-s)}%
q(k,s)~ds~,\label{pm4}%
\end{align}
with $\tilde{\omega}_{-}^{\ast}$ and $\tilde{u}_{-}^{\ast}$ to be chosen later
on. The integral equations (\ref{pm1})-(\ref{pm4}) are identical to the ones
discussed in \cite{pw001}, \cite{pw002}. There, by restricting to symmetric
flows, $q$ was an odd function of $k$ and therefore, in a space of
continuously differentiable functions, the division by $k$ that appears in
(\ref{pm3}) and (\ref{pm4}) was compensated by a factor of $k$ coming from
$q$, so that $u_{+}$ and $u_{-}$ were continuous functions of $k$. For
non-symmetric functions $q$ this strategy does not work anymore. As mentioned
in the introduction, it is replaced in what follows by a more detailed
analysis of the nonlinearity $q$, and by using the invariance properties of
the equations which make that the singular terms compensate each other in the
physically relevant functions $\omega$, $\eta$, $u$ and $v$ as given by
(\ref{uv}). In fact, after substitution of the integral equations
(\ref{pm1})-(\ref{pm4}) into the change of coordinates (\ref{uv}) we get, with
(\ref{qq}), and after integrating by parts the time derivative acting on
$q_{0}$, the following integral equations for $\omega$, $u$ and $v $:%
\begin{align}
\omega(k,t)  & =\left(  \tilde{\omega}_{-}^{\ast}(k)+\frac{1}{\Lambda_{0}%
}q_{0}(k,1)\right)  e^{\Lambda_{-}(t-1)}\nonumber\\
& +\frac{1}{\Lambda_{0}}\int_{1}^{t}e^{\Lambda_{-}(t-s)}ikq_{1}%
(k,s)~ds\nonumber\\
& +\frac{1}{\Lambda_{0}}\int_{t}^{\infty}e^{\Lambda_{+}(t-s)}ikq_{1}%
(k,s)~ds\nonumber\\
& -\frac{\Lambda_{-}}{\Lambda_{0}}\int_{1}^{t}e^{\Lambda_{-}(t-s)}%
q_{0}(k,s)~ds\nonumber\\
& -\frac{\Lambda_{+}}{\Lambda_{0}}\int_{t}^{\infty}e^{\Lambda_{+}(t-s)}%
q_{0}(k,s)~ds~,\label{omega_t}\\
u(k,t)  & =-\frac{i}{k}\Lambda_{+}\left(  \tilde{\omega}_{-}^{\ast}%
(k)+\frac{1}{\Lambda_{0}}q_{0}(k,1)\right)  e^{\Lambda_{-}(t-1)}\nonumber\\
& +\left(  \tilde{u}_{-}^{\ast}(k)+\frac{1}{2}\frac{i}{k}q_{0}(k,1)\right)
e^{-\left\vert k\right\vert (t-1)}\nonumber\\
& +\frac{\Lambda_{+}}{\Lambda_{0}}\int_{1}^{t}e^{\Lambda_{-}(t-s)}%
q_{1}(k,s)~ds\nonumber\\
& -\frac{1}{2}\int_{1}^{t}e^{-\left\vert k\right\vert (t-s)}q_{1}%
(k,s)~ds\nonumber\\
& +\frac{1}{2}\int_{t}^{\infty}e^{\left\vert k\right\vert (t-s)}%
q_{1}(k,s)~ds\nonumber\\
& +\frac{\Lambda_{-}}{\Lambda_{0}}\int_{t}^{\infty}e^{\Lambda_{+}(t-s)}%
q_{1}(k,s)~ds\nonumber\\
& +\frac{1}{2}i\sigma(k)\int_{1}^{t}e^{-\left\vert k\right\vert (t-s)}%
q_{0}(k,s)~ds\nonumber\\
& +\frac{1}{2}i\sigma(k)\int_{t}^{\infty}e^{\left\vert k\right\vert
(t-s)}q_{0}(k,s)~ds\nonumber\\
& -\frac{ik}{\Lambda_{0}}\int_{1}^{t}e^{\Lambda_{-}(t-s)}q_{0}%
(k,s)~ds\nonumber\\
& -\frac{ik}{\Lambda_{0}}\int_{t}^{\infty}e^{\Lambda_{+}(t-s)}q_{0}%
(k,s)~ds~,\label{u_t}\\
v(k,t)  & =\omega(k,t)\nonumber\\
& +i\sigma(k)\left(  \tilde{u}_{-}^{\ast}(k)+\frac{1}{2}\frac{i}{k}%
q_{0}(k,1)\right)  e^{-\left\vert k\right\vert (t-1)}\nonumber\\
& -\frac{1}{2}i\sigma(k)\int_{1}^{t}e^{-\left\vert k\right\vert (t-s)}%
q_{1}(k,s)~ds\nonumber\\
& -\frac{1}{2}i\sigma(k)\int_{t}^{\infty}e^{\left\vert k\right\vert
(t-s)}q_{1}(k,s)~ds\nonumber\\
& -\frac{1}{2}\int_{1}^{t}e^{-\left\vert k\right\vert (t-s)}q_{0}%
(k,s)~ds\nonumber\\
& +\frac{1}{2}\int_{t}^{\infty}e^{\left\vert k\right\vert (t-s)}%
q_{0}(k,s)~ds~,\label{v_t}%
\end{align}
with $q_{0}$ and $q_{1}$ given by (\ref{q0o}) and (\ref{q1o}), respectively.
Note that the function $\eta$ does not need to be constructed since it does
not appear in the nonlinearities $q_{0}$ and $q_{1}$.

A closer look at (\ref{omega_t})-(\ref{v_t}) reveals, that the problem
concerning the division by $k$ in the equations (\ref{pm1})-(\ref{pm4}) has
not disappeared. However, in this new representation, the invariance
properties of the equations have become manifest, and we see that the problem
can be eliminated by a proper choice of initial conditions, \textit{i.e.},
$\omega$, $u$, and $v$ are either regular or singular for all times. In
particular, as we will see, if we set%
\begin{align}
\tilde{\omega}_{-}^{\ast}(k)  & =-ik\omega_{-}^{\ast}(k)-\frac{1}{\Lambda_{0}%
}q_{0}(k,1)~,\label{wmt}\\
\tilde{u}_{-}^{\ast}(k)  & =u_{-}^{\ast}(k)-\frac{1}{2}\frac{i}{k}%
q_{0}(k,1)~,\label{umt}%
\end{align}
with%
\begin{equation}
u_{-}^{\ast}(k)=u_{-,1}^{\ast}(k)-i\sigma(k)u_{-,2}^{\ast}(k)~,\label{um}%
\end{equation}
and with $\omega_{-}^{\ast}$, $u_{-,1}^{\ast}$, and $u_{-,2}^{\ast}$ smooth,
then $\omega$ and $q$ are smooth, and $u$ and $v$ are smooth modulo
discontinuities at $k=0$. This corresponds to choosing initial conditions
exactly as singular as dictated by the nonlinearity. We expect this choice to
be general enough to cover all cases of stationary exterior flows.

Below, we will prove existence of solutions to (\ref{omega_t})-(\ref{v_t}) for
certain classes of continuous complex valued functions $\omega_{-}^{\ast}$,
$u_{-,1}^{\ast}$, and $u_{-,2}^{\ast}$. Once the existence of solutions has
been established, we will restrict attention to even, real valued functions
$u_{-,1}^{\ast}$, and $u_{-,2}^{\ast}$, and to complex valued functions
$\omega_{-}^{\ast}$ of the form%
\begin{equation}
\omega_{-}^{\ast}(k)=\omega_{-,1}^{\ast}(k)+i\omega_{-,2}^{\ast}%
(k)~,\label{wm}%
\end{equation}
with $\omega_{-,1}^{\ast}$ and $\omega_{-,2}^{\ast}$ real valued, even and odd
functions of $k$, respectively. This corresponds to the restriction to real
valued solutions of (\ref{ds}).

It turns out that the decomposition of the nonlinearity $q$ into $q_{0}$ and
$q_{1}$ is not detailed enough to prove the existence of a solution to
(\ref{omega_t})-(\ref{v_t}). To overcome this problem we split $\omega$ and
$u$ into a \textquotedblleft dominant part\textquotedblright\ and a
\textquotedblleft remainder.\textquotedblright\ Namely, we set%
\begin{align}
\omega(k,t)  & =\omega_{0}(k,t)+\omega_{1}(k,t)~,\label{omega}\\
u(k,t)  & =u_{0}(k,t)+u_{1}(k,t)~,\label{u}%
\end{align}
with $\omega$ and $u$ given by (\ref{omega_t}) and (\ref{u_t}), and with%
\begin{align}
u_{0}(k,t)  & =-\omega_{-}^{\ast}(k)e^{\Lambda_{-}(t-1)}+\frac{1}{\Lambda_{0}%
}\int_{1}^{t}e^{\Lambda_{-}(t-s)}q_{1}(k,s)~ds~,\label{u0}\\
\omega_{0}(k,t)  & =ik~u_{0}(k,t)~.\label{omega0}%
\end{align}
This decomposition allows us to split the function $q_{0}$ into a term which
is \textquotedblleft large\textquotedblright, but zero at $k=0$, and a
remaining term which is \textquotedblleft small\textquotedblright\ (see
Section 4 for the definition of \textquotedblleft large\textquotedblright\ and
\textquotedblleft small\textquotedblright). Namely, using (\ref{omega0}) we
find that%
\begin{align*}
\left(  u_{0}\ast\omega_{0}\right)  (k,t)  & =\left(  u_{0}\ast(iku_{0}%
)\right)  (k,t)\\
& =\frac{1}{2}ik~(u_{0}\ast u_{0})(k,t)~,
\end{align*}
and therefore, using the definition (\ref{q0o}) of $q_{0}$, we find that%
\begin{equation}
q_{0}(k,t)=q_{0,0}(k,t)-ik~q_{0,1}(k,t)~,\label{q0}%
\end{equation}
with%
\begin{align}
q_{0,0}(k,t)  & =\frac{1}{2\pi}\left(  u_{0}\ast\omega_{1}+u_{1}\ast\omega
_{0}+u_{1}\ast\omega_{1}\right)  (k,t)~,\label{q00}\\
q_{0,1}(k,t)  & =-\frac{1}{4\pi}(u_{0}\ast u_{0})(k,t)~.\label{q01}%
\end{align}
After some rearrangement, and using (\ref{wmt})-(\ref{um}), we find for
$\omega_{1}$ and $u_{1}$ instead of the equations (\ref{omega}) and (\ref{u})
the following explicit expressions, which we will use in the sequel:%
\begin{align}
\omega_{1}(k,t)  & =\frac{1}{\Lambda_{0}}\int_{t}^{\infty}e^{\Lambda_{+}%
(t-s)}ikq_{1}(k,s)~ds\nonumber\\
& -\frac{\Lambda_{+}}{\Lambda_{0}}\int_{t}^{\infty}e^{\Lambda_{+}(t-s)}%
q_{0}(k,s)~ds\nonumber\\
& -\frac{\Lambda_{-}}{\Lambda_{0}}\int_{1}^{t}e^{\Lambda_{-}(t-s)}%
q_{0}(k,s)~ds~,\label{omega1}%
\end{align}
and%
\begin{align}
u_{1}(k,t)  & =\left(  u_{-,1}^{\ast}(k)-i\sigma(k)u_{-,2}^{\ast}(k)\right)
e^{-\left\vert k\right\vert (t-1)}\nonumber\\
& +\omega_{-}^{\ast}(k)\Lambda_{-}e^{\Lambda_{-}(t-1)}\nonumber\\
& -\frac{\Lambda_{-}}{\Lambda_{0}}\int_{1}^{t}e^{\Lambda_{-}(t-s)}%
q_{1}(k,s)~ds\nonumber\\
& -\frac{1}{2}\int_{1}^{t}e^{-\left\vert k\right\vert (t-s)}q_{1}%
(k,s)~ds\nonumber\\
& +\frac{1}{2}\int_{t}^{\infty}e^{\left\vert k\right\vert (t-s)}%
q_{1}(k,s)~ds\nonumber\\
& +\frac{\Lambda_{-}}{\Lambda_{0}}\int_{t}^{\infty}e^{\Lambda_{+}(t-s)}%
q_{1}(k,s)~ds\nonumber\\
& -\frac{ik}{\Lambda_{0}}\int_{1}^{t}e^{\Lambda_{-}(t-s)}q_{0}%
(k,s)~ds\nonumber\\
& +\frac{1}{2}i\sigma(k)\int_{1}^{t}e^{-\left\vert k\right\vert (t-s)}%
q_{0}(k,s)~ds\nonumber\\
& +\frac{1}{2}i\sigma(k)\int_{t}^{\infty}e^{\left\vert k\right\vert
(t-s)}q_{0}(k,s)~ds\nonumber\\
& -\frac{ik}{\Lambda_{0}}\int_{t}^{\infty}e^{\Lambda_{+}(t-s)}q_{0}%
(k,s)~ds~.\label{u1}%
\end{align}


\section{Function spaces}

In order to prove the existence of a solution for (\ref{omega_t})-(\ref{v_t})
we will apply, for fixed $\omega_{-}^{\ast}$ and fixed $u_{-,1}^{\ast}$,
$u_{-,2}^{\ast}$, the contraction mapping principle to the map $(\tilde
{q}_{0,0},\tilde{q}_{0,1},\tilde{q}_{1})=\mathcal{N}(q_{0,0},q_{0,1},q_{1})$
that is formally defined by computing first $q_{0}$ from $q_{0,0}$ and
$q_{0,1}$ using (\ref{q0}), then $\omega_{0}$, $\omega_{1}$, $\omega$, $u_{0}%
$, $u_{1}$, $u$ and $v$ using (\ref{omega0}), (\ref{omega1}), (\ref{omega}),
(\ref{u0}), (\ref{u1}), (\ref{u}), (\ref{v_t}) and (\ref{umt}) and then
$q_{0,0}$, $q_{0,1}$ and $q_{1}$ by using (\ref{q00}), (\ref{q01}) and
(\ref{q1o}). We now define the functions spaces that will be used below:

Let $\alpha$, $p\geq0$ and
\begin{equation}
\mu_{\alpha}^{p}(k,t)=\frac{1}{1+\left(  \left\vert k\right\vert t^{p}\right)
^{\alpha}}~.\label{mu}%
\end{equation}
Let furthermore%
\begin{align*}
\mu_{\alpha}(k,t)  & =\mu_{a}^{1/2}(k,t)~,\\
\bar{\mu}_{\alpha}(k,t)  & =\mu_{a}^{1}(k,t)~.
\end{align*}
We then consider, for fixed $\alpha\geq0$, the Banach space $\mathcal{V}%
_{\alpha}$ of functions $f\in\mathcal{C}(\mathbf{R,C})$ (continuous functions
from $\mathbf{R}$ to $\mathbf{C}$), equipped with the norm
\[
\left\Vert f\right\Vert _{\alpha}=\sup_{k\in\mathbf{R}}\frac{\left\vert
f(k)\right\vert }{\mu_{\alpha}(k,1)}~,
\]
and, for fixed $\alpha$, $\beta\geq0$, the Banach space $\mathcal{B}%
_{\alpha,\beta}$ of functions $f\in\mathcal{C}([1,\infty),\mathcal{V}_{\alpha
}),$ equipped with the norm
\[
\left\Vert f\right\Vert _{\alpha,\beta}=\sup_{t\geq1}t^{\beta}||f(t^{-1/2}%
.~,t)||_{\alpha}~.
\]


Finally, we define the Banach space $\mathcal{B}_{\alpha}$,%
\[
\mathcal{B}_{\alpha}=\mathcal{B}_{\alpha,3/2}\oplus\mathcal{B}_{\alpha
+1,1/2}\oplus\mathcal{B}_{\alpha,3/2}~,
\]
equipped with the norm%
\[
\left\Vert (\rho_{0},\rho_{1},\rho_{2})\right\Vert _{\alpha}=\left\Vert
\rho_{0}\right\Vert _{\alpha,3/2}+\left\Vert \rho_{1}\right\Vert
_{\alpha+1,1/2}+\left\Vert \rho_{2}\right\Vert _{\alpha,3/2}~.
\]


\begin{theorem}
\label{theorem1}Let $\alpha>1$. Let $u_{-,1}^{\ast}$, $u_{-,2}^{\ast}$,
$\omega_{-}^{\ast}\in\mathcal{V}_{\alpha+1}$, and let $\varepsilon
_{0}=\left\Vert u_{-,1}^{\ast}\right\Vert _{\alpha+1}+\left\Vert u_{-,2}%
^{\ast}\right\Vert _{\alpha+1}+\left\Vert \omega_{-}^{\ast}\right\Vert
_{\alpha+1}$. Then, $\mathcal{N}$ is well defined as a map from $\mathcal{B}%
_{\alpha}$ to $\mathcal{B}_{\alpha}$ and contracts, for $\varepsilon_{0}$
sufficiently small, the ball $B_{\alpha}(\varepsilon_{0})=\left\{  \rho
\in\mathcal{B}_{\alpha}\left\vert {}\right.  \left\Vert \rho\right\Vert
_{\alpha}\leq\varepsilon_{0}\right\}  $ into itself.
\end{theorem}

Theorem \ref{theorem1} implies that for $\varepsilon_{0}$ small enough
$\mathcal{N}$ has a unique fixed point in $B_{\alpha}(\varepsilon_{0})$,
\textit{i.e.}, the integral equations (\ref{omega_t})-(\ref{v_t}) have a solution.

\section{Proof of Theorem \ref{theorem1}}

The proof is organized as follows: we first prove that $\mathcal{N}$ is well
defined and maps, for small enough initial conditions $\omega_{-}^{\ast}$ and
$u_{-}^{\ast}$, a ball in $\mathcal{B}_{\alpha}$ into itself. Then, we show
that $\mathcal{N}$ is a contraction on this ball.

Let $\varepsilon_{0}$ be as in Theorem \ref{theorem1}. Throughout all proofs
we then denote by $\varepsilon$ a constant multiple of $\varepsilon_{0}$,
\textit{i.e.}, $\varepsilon=\mathrm{const.}~\varepsilon_{0}$ with a constant
that may be different from instance to instance.

\subsection{$\mathcal{N}$ is well defined}

We first prove bounds on $\omega_{0}$, $\omega_{1}$, $u_{0}$, $u_{1}$ and $v$:

\begin{proposition}
\label{pp11}Let $\alpha>0$. Let $u_{-,1}^{\ast}$, $u_{-,2}^{\ast}$,
$\omega_{-}^{\ast}\in\mathcal{V}_{\alpha+1}$, with $\varepsilon_{0}=\left\Vert
u_{-,1}^{\ast}\right\Vert _{\alpha+1}+\left\Vert u_{-,2}^{\ast}\right\Vert
_{\alpha+1}+\left\Vert \omega_{-}^{\ast}\right\Vert _{\alpha+1}$, and let
$(q_{0,0},q_{0,1},q_{1})\in B_{\alpha}(\varepsilon)$. Then, $\omega_{0}$,
$\omega_{1}$ and $u_{0}$ as defined by (\ref{omega0}), (\ref{omega1}) and
(\ref{u0}) are continuous functions from $\mathbf{R}\times\lbrack1,\infty)$ to
$\mathbf{C}$, and $u_{1}$ and $v$ as defined by (\ref{u1}) and (\ref{v_t}) are
of the form%
\begin{align}
u_{1}(k,t)  & =u_{1,E}(k,t)+i\sigma(k)u_{1,O}(k,t)~,\label{dec1}\\
v(k,t)  & =v_{E}(k,t)+i\sigma(k)v_{O}(k,t)~\label{dec2}%
\end{align}
with $u_{1,E}$, $u_{1,O}$, $v_{E}$ and $v_{O}$ continuous functions from
$\mathbf{R}\times\lbrack1,\infty)$ to $\mathbf{C}$. Furthermore, we have the
bounds
\begin{align}
\left\vert \omega_{0}(k,t)\right\vert  & \leq\frac{\varepsilon}{t^{1/2}}%
\mu_{\alpha}(k,t)~,\label{bpm1}\\
\left\vert \omega_{1}(k,t)\right\vert  & \leq\frac{\varepsilon}{t}\mu_{\alpha
}(k,t)~,\label{bpm2}\\
\left\vert u_{0}(k,t)\right\vert  & \leq\varepsilon\mu_{\alpha+1}%
(k,t)~,\label{bpm3}\\
\left\vert u_{1}(k,t)\right\vert  & \leq\varepsilon\bar{\mu}_{\alpha
+1}(k,t)+\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}(k,t)~,\label{bpm4}\\
\left\vert v(k,t)\right\vert  & \leq\varepsilon\bar{\mu}_{\alpha+1}%
(k,t)+\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}(k,t)~.\label{bpm5}%
\end{align}
uniformly in $k\in\mathbf{R}$ and $t\geq1$.
\end{proposition}

See Appendix I for a proof.

\bigskip

Now we prove bounds on $q_{0,0}$, $q_{0,1}$ and $q_{1}$:

\begin{proposition}
\label{pp12}Let $\alpha>1$. Let $\omega_{0}$, $\omega_{1}$ and $u_{0}$ be
continuous functions from $\mathbf{R}\times\lbrack1,\infty)$ to $\mathbf{C}$
satisfying the bounds (\ref{bpm1})-(\ref{bpm3}), and let $u_{1}$ and $v$ be
continuous functions from $\mathbf{R}\setminus\{0\}\times\lbrack1,\infty)$ to
$\mathbf{C}$, satisfying the bounds (\ref{bpm4}) and (\ref{bpm5}),
respectively. Then, $q_{0,0}$, $q_{0,1}$ and $q_{1}$ as defined by
(\ref{q00}), (\ref{q01}) and (\ref{q1o}) are continuous functions from
$\mathbf{R}\times\lbrack0,\infty)$ to $\mathbf{C}$, and we have the bounds
\begin{align}
\left\vert q_{0,0}(k,t)\right\vert  & \leq\frac{\varepsilon^{2}}{t^{3/2}}%
\mu_{a}(k,t)~,\label{nqb1}\\
\left\vert q_{0,1}(k,t)\right\vert  & \leq\frac{\varepsilon^{2}}{t^{1/2}}%
\mu_{a+1}(k,t)~,\label{nqb2}\\
\left\vert q_{1}(k,t)\right\vert  & \leq\frac{\varepsilon^{2}}{t^{3/2}}\mu
_{a}(k,t)~,\label{nqb3}%
\end{align}
uniformly in $k\in\mathbf{R}$ and $t\geq1$, and therefore $\left\Vert
(q_{0,0},q_{0,1},q_{1})\right\Vert _{\alpha}\leq\varepsilon^{2}$.
\end{proposition}

See Appendix II for a proof.

\bigskip

Proposition \ref{pp11} together with Proposition \ref{pp12} imply that, for
$\rho\in B_{\alpha}(\varepsilon)$, $\left\Vert \mathcal{N}(\rho)\right\Vert
_{\alpha}\leq\varepsilon^{2}$. Therefore, $\mathcal{N}$ is well defined as a
map from $\mathcal{B}_{\alpha}$ to $\mathcal{B}_{\alpha}$. Furthermore, since
$\varepsilon^{2}=\mathrm{const.}$ $\varepsilon_{0}^{2} $, it follows that
$\mathcal{N}$ maps $B_{\alpha}(\varepsilon_{0}) $ into itself for
$\varepsilon_{0}$ small enough.

\subsection{$\mathcal{N}$ is Lipschitz}

In order to complete the proof of Theorem \ref{theorem1} it remains to be
shown that $\mathcal{N}$ is Lipschitz:

\begin{proposition}
\label{prop12}Let $\alpha>1$. Let $u_{-,1}^{\ast}$, $u_{-,2}^{\ast}$,
$\omega_{-}^{\ast}\in\mathcal{V}_{\alpha+1}$, with $\varepsilon_{0}=\left\Vert
u_{-,1}^{\ast}\right\Vert _{\alpha+1}+\left\Vert u_{-,2}^{\ast}\right\Vert
_{\alpha+1}+\left\Vert \omega_{-}^{\ast}\right\Vert _{\alpha+1}$, and let
$\rho$, $\tilde{\rho}\in B_{\alpha}(\varepsilon_{0})$. Then
\begin{equation}
\left\Vert \mathcal{N}(\rho)-\mathcal{N}(\tilde{\rho})\right\Vert _{\alpha
}\leq\varepsilon\left\Vert \rho-\tilde{\rho}\right\Vert _{\alpha
}~.\label{lipschitz}%
\end{equation}

\end{proposition}

See Appendix III for a proof.

\bigskip

Proposition \ref{pp11} together with Proposition \ref{pp12} show that, for
$\alpha>1$, $\mathcal{N}$ maps the ball $B_{\alpha}(\varepsilon_{0})$ into
itself for $\varepsilon_{0}$ small enough, and Proposition \ref{prop12}
therefore shows that $\mathcal{N}$ is a contraction of $B_{\alpha}%
(\varepsilon_{0})$ into itself for $\varepsilon_{0}$ small enough. This
completes the proof of Theorem \ref{theorem1}.

\section{Invariant quantities}

We now restrict attention to real valued, even functions $u_{-,1}^{\ast}$, and
$u_{-,2}^{\ast}$, and to complex valued functions $\omega_{-}^{\ast}$ of the
form (\ref{wm}), with $\omega_{-,1}^{\ast}$ and $\omega_{-,2}^{\ast}$ real
valued, even and odd functions of $k$, respectively.

\begin{proposition}
In the limit $k\rightarrow0_{\pm}$ the equations (\ref{u}) and (\ref{v_t})
reduce to%
\begin{align}
u(0_{\pm},t)  & =-\omega_{-}^{\ast}(0)+\frac{1}{2}\int_{1}^{\infty}%
q_{1}(0,s)~ds+u_{-,1}^{\ast}(0)\nonumber\\
& \pm i\left(  -u_{-,2}^{\ast}(0)+\frac{1}{2}\int_{1}^{\infty}q_{0}%
(0,s)~ds\right)  ~,\label{upm}\\
v(0_{\pm},t)  & =u_{-,2}^{\ast}(0)-\frac{1}{2}\int_{1}^{\infty}q_{0}%
(0,s)~ds\nonumber\\
& +\int_{t}^{\infty}(1-e^{t-s})q_{0}(0,s)~ds\nonumber\\
& \pm i\left(  u_{-,1}^{\ast}(0)-\frac{1}{2}\int_{1}^{\infty}q_{1}%
(0,s)~ds\right)  ~,\label{vpm}%
\end{align}


\begin{proof}
This follows immediately using the fact that by Proposition \ref{pp11} the
functions $u$ and $v$ are continuous on $[0,\infty)$ and $(-\infty,0]$,
respectively, for all $t\geq1$.
\end{proof}
\end{proposition}

From (\ref{upm}) and (\ref{vpm}) we see that the three (real) quantities $b$,
$c$ and $d$,
\begin{align*}
b  & =\frac{u(0_{+},t)-u(0_{-},t)}{2i}~,\\
d  & =\frac{v(0_{+},t)-v(0_{-},t)}{2i}~,\\
c  & =\frac{u(0_{+},t)+u(0_{-},t)}{2}-d~,
\end{align*}
are independent of $t\geq1$. Explicitly, we have%
\begin{align}
b  & =-u_{-,2}^{\ast}(0)+\frac{1}{2}\int_{1}^{\infty}q_{0}(0,s)~ds~,\label{b}%
\\
d  & =u_{-,1}^{\ast}(0)-\frac{1}{2}\int_{1}^{\infty}q_{1}(0,s)~ds~,\label{d}\\
c  & =-\omega_{-}^{\ast}(0)+\int_{1}^{\infty}q_{1}(0,s)~ds~.\label{c}%
\end{align}
We also note that the quantity $\varphi$,
\begin{equation}
\varphi=2d+c=2u_{-,1}^{\ast}(0)-\omega_{-}^{\ast}(0)\label{zero_flux}%
\end{equation}
is directly given in terms of the initial conditions (see \cite{pw001},
\cite{pw002} and \cite{abc} for the physical interpretation of $\varphi$).

\section{Asymptotic behavior}

The following theorem provides the leading order behavior of solutions whose
existence has been shown in Theorem \ref{theorem1}. We again restrict
attention to even, real valued functions $u_{-,1}^{\ast}$, and $u_{-,2}^{\ast
}$, and to complex valued functions $\omega_{-}^{\ast}$ of the form
(\ref{wm}), with $\omega_{-,1}^{\ast}$ and $\omega_{-,2}^{\ast}$ real valued,
even and odd functions of $k$, respectively.

\begin{theorem}
\label{th4}Let $\alpha>1$. Let $u_{-,1}^{\ast}$, $u_{-,2}^{\ast}$, $\omega
_{-}^{\ast}\in\mathcal{V}_{\alpha+1}$, with $\varepsilon_{0}=\left\Vert
u_{-,1}^{\ast}\right\Vert _{\alpha+1}+\left\Vert u_{-,2}^{\ast}\right\Vert
_{\alpha+1}+\left\Vert \omega_{-}^{\ast}\right\Vert _{\alpha+1}$ sufficiently
small. Then, the equations (\ref{omega_t})-(\ref{v_t}) have a solution and%
\begin{align}
\lim_{t\rightarrow\infty}t^{1/2}\int_{\mathbf{R}}\left\vert u(k,t)-u_{as}%
(k,t)\right\vert ~dk~  & =0~,\label{uas}\\
\lim_{t\rightarrow\infty}t\int_{\mathbf{R}}\left\vert v(k,t)-v_{as}%
(k,t)\right\vert ~dk~  & =0~,\label{vas}%
\end{align}
where
\begin{align*}
u_{as}(k,t)  & =c~e^{-k^{2}t}+d~e^{-\left\vert k\right\vert t}+b~i\sigma
(k)e^{-\left\vert k\right\vert t}~,\\
v_{as}(k,t)  & =c~ike^{-k^{2}t}+d~i\sigma(k)e^{-\left\vert k\right\vert
t}-b~e^{-\left\vert k\right\vert t}~,
\end{align*}
with $b$, $c$ and $d$ as defined in (\ref{b}), (\ref{d}) and (\ref{c}).
\end{theorem}

The existence of a solution follows from Theorem \ref{theorem1}. A proof of
(\ref{uas}) and (\ref{vas}) can be found in Appendix IV.

\section{Proof of Theorem \ref{theorem0}}

We again restrict attention to even, real valued functions $u_{-,1}^{\ast}$,
and $u_{-,2}^{\ast}$, and to complex valued functions $\omega_{-}^{\ast} $ of
the form (\ref{wm}), with $\omega_{-,1}^{\ast}$ and $\omega_{-,2}^{\ast}$ real
valued, even and odd functions of $k$, respectively. For $\alpha>1$ we have
proved in Section 5 the existence of a solution of the equations
(\ref{omega_t})-(\ref{v_t}) satisfying (to avoid confusion we now write the
hats for the Fourier transforms)
\begin{align}
\left\vert \hat{u}(k,t)\right\vert  & \leq\varepsilon\mu_{\alpha
}(k,t)~,\label{uubound}\\
\left\vert \hat{v}(k,t)\right\vert  & \leq\frac{\varepsilon}{t^{1/2}}%
\mu_{\alpha}(k,t)+\varepsilon\bar{\mu}_{\alpha+1}(k,t)~.\label{uvbound}%
\end{align}
Since, for $\alpha>1$, the real and imaginary parts of the functions
$k\mapsto\hat{u}(k,t)$ and $k\mapsto\hat{v}(k,t)$ are respectively even and
odd functions in $L^{1}(\mathbf{R},dk)$ for all $t\geq1$, their Fourier
transforms%
\begin{align*}
u(x,y)  & =\int_{\mathbf{R}}e^{-iky}\hat{u}(k,x)~dk~,\\
v(x,y)  & =\int_{\mathbf{R}}e^{-iky}\hat{v}(k,x)~dk~,
\end{align*}
are by the Riemann-Lebesgue lemma real valued continuous functions of $y$ and
vanish as $\left\vert y\right\vert \rightarrow$ $\infty$ for each $x\geq1$.
Moreover, using (\ref{uubound}) and (\ref{uvbound}), we find that
\begin{align}
\sup_{y\in\mathbf{R}}\left\vert u(x,y)\right\vert  & \leq\frac{\varepsilon
}{\left\vert x\right\vert ^{1/2}}~,\label{ubound1}\\
\sup_{y\in\mathbf{R}}\left\vert v(x,y)\right\vert  & \leq\frac{\varepsilon
}{\left\vert x\right\vert }~.\label{vbound1}%
\end{align}
As a consequence, $u$ and $v$ converge to zero whenever $\left\vert
x\right\vert +\left\vert y\right\vert \rightarrow\infty$ in $\Omega$ (see
Section 5 of \cite{pw001} for details), and satisfy therefore not only
(\ref{ww}) but also the boundary conditions (\ref{bc001}), (\ref{bc002}). The
reconstruction of the pressure from $u$ and $v$ is standard. For $\alpha>3$
second derivatives of $u$ and $v$ are continuous in direct space, and one
easily verifies using the definitions that the triple $(u,v,p)$ satisfies the
Navier-Stokes equations (\ref{navier}). The set $\mathcal{S}$ in Theorem
\ref{theorem0} is by definition the set of all vector fields $(u,v)$ obtained
this way, restricted to $\Sigma$. Finally, equations (\ref{th2})-(\ref{ccc})
are a direct consequence of Theorem \ref{th4}. This completes the proof of
Theorem \ref{theorem0}.

\section{Appendix I}

In this appendix we give a proof of Proposition \ref{pp11}. We first prove the
continuity, then the bounds.

\bigskip

The continuity of the functions $u_{0}$ and $\omega_{0}$ is elementary.
Similarly, using that $\Lambda_{+}(k)\geq1$, the continuity of the function
$\omega_{1}$ is elementary, since the improper integrals in (\ref{omega1})
converge uniformly in $k$. Next, we note that $u_{1}$ and $v$ as given by
(\ref{u1}) and (\ref{v_t}) are explicitly of the form (\ref{dec1}),
(\ref{dec2}). The continuity of the functions $u_{1,E}$, $u_{1,O}$, $v_{E}$
and $v_{O}$ is again elementary except for the improper integrals involving
the function $e^{\left\vert k\right\vert (t-s)}$. There are two cases of such
integrals; those involving $q_{1}$ and those involving $q_{0}$. Since
$\left\vert q_{1}(k,s)\right\vert \leq\varepsilon/s^{3/2}$ and since
$1/s^{3/2}$ is integrable at infinity, the continuity follows in these cases.
Similarly, since $\left\vert q_{0,0}(k,s)\right\vert \leq\varepsilon/s^{3/2}$,
the contribution of $q_{0,0}$ to the integrals involving $q_{0}$ defines
continuous functions. This leaves us with the case of improper integrals
involving $e^{\left\vert k\right\vert (t-s)}$ and $q_{0,1}$. For these cases
we have the inequality%
\begin{align}
\left\vert \frac{1}{2}i\sigma(k)\int_{t}^{\infty}e^{\left\vert k\right\vert
(t-s)}\left(  -ikq_{0,1}(k,s)\right)  ~ds\right\vert  & \leq\int_{t}^{\infty
}e^{\left\vert k\right\vert (t-s)}\frac{\varepsilon}{s^{1/2}}\left\vert
k\right\vert \mu_{a+1}(k,s)~ds\nonumber\\
& \leq\varepsilon\mu_{a+1}(k,t)\int_{t}^{\infty}e^{\left\vert k\right\vert
(t-s)}\left\vert k\right\vert ~\frac{ds}{s^{1/2}}\nonumber\\
& \leq\varepsilon\mu_{a+1}(k,t)\left\vert k\right\vert ^{1/2}e^{\left\vert
k\right\vert t}\int_{\left\vert k\right\vert t}^{\infty}e^{-\sigma}%
\frac{d\sigma}{\sigma^{1/2}}\nonumber\\
& \leq\varepsilon\mu_{a+1}(k,t)\left\vert k\right\vert ^{1/2}e^{\left\vert
k\right\vert t}(1-\operatorname{erf}(\left\vert k\right\vert t))\nonumber\\
& \leq\varepsilon\left\vert k\right\vert ^{1/2}\bar{\mu}_{1/2}(k,t)\mu
_{a+1}(k,t)~,\label{ccbound}%
\end{align}
with $\operatorname{erf}$ the error function. This shows continuity, since
$\lim_{k\rightarrow0}\varepsilon\left\vert k\right\vert ^{1/2}\bar{\mu}%
_{1/2}(k,t)\mu_{a+1}(k,t)=0$ for $t\geq1$. This completes the proof of the continuity.

\bigskip

We next prove the bounds (\ref{bpm1})-(\ref{bpm5}). We start by proving an
inequality which will be routinely used below:

\subsection{Main technical Lemma}

\begin{proposition}
\label{basic}Let $\alpha^{\prime}\geq\beta^{\prime}\geq\gamma^{\prime}\geq0$
and $\mu>0$. Then, we have the bound%
\begin{equation}
\frac{1}{1+\left\vert k\right\vert ^{\alpha^{\prime}}}e^{\mu\Lambda_{-}\left(
t-1\right)  }\left\vert \Lambda_{-}\right\vert ^{\beta^{\prime}}\left(
\dfrac{t-1}{t}\right)  ^{\gamma^{\prime}}\leq\mathrm{const.}~\frac{1}%
{t^{\beta^{\prime}}}\frac{1}{1+\left(  \left\vert k\right\vert t^{1/2}\right)
^{\alpha^{\prime}-\beta^{\prime}+\gamma^{\prime}}}~,\label{decay1}%
\end{equation}
uniformly in $k\in\mathbf{R}$ and $t\geq1$. Similarly, for positive
$\alpha^{\prime}$, $\beta^{\prime}$, $\gamma^{\prime}$with $\alpha^{\prime
}-\beta^{\prime}+\gamma^{\prime}\geq0$ and $\mu>0$ we have the bound%
\begin{equation}
\frac{1}{1+\left\vert k\right\vert ^{\alpha^{\prime}}}e^{-\mu\left\vert
k\right\vert \left(  t-1\right)  }\left\vert k\right\vert ^{\beta^{\prime}%
}\left(  \dfrac{t-1}{t}\right)  ^{\gamma^{\prime}}\leq\mathrm{const.}~\frac
{1}{t^{\beta^{\prime}}}\frac{1}{1+\left(  \left\vert k\right\vert t\right)
^{\alpha^{\prime}-\beta^{\prime}+\gamma^{\prime}}}~,\label{decay2}%
\end{equation}
uniformly in $k\in\mathbf{R}$ and $t\geq1$.
\end{proposition}

\begin{proof}
We first prove (\ref{decay1}). For $1\leq t\leq2$ we have that%
\begin{align*}
\frac{1}{1+\left\vert k\right\vert ^{\alpha^{\prime}}}e^{\mu\Lambda_{-}%
(t-1)}\left\vert \Lambda_{-}\right\vert ^{\beta^{\prime}}\left(  \dfrac
{t-1}{t}\right)  ^{\gamma^{\prime}} &  \leq\mathrm{const.}~\frac
{1}{1+\left\vert k\right\vert ^{\alpha^{\prime}}}e^{\mu\Lambda_{-}%
(t-1)}\left\vert \Lambda_{-}\left(  t-1\right)  \right\vert ^{\gamma^{\prime}%
}~\left\vert \Lambda_{-}\right\vert ^{\beta^{\prime}-\gamma^{\prime}}\\
&  \leq\mathrm{const.}~\frac{1}{1+\left\vert k\right\vert ^{\alpha^{\prime}}%
}~\left\vert \Lambda_{-}\right\vert ^{\beta^{\prime}-\gamma^{\prime}}\\
&  \leq\mathrm{const.}~\frac{1}{1+\left\vert k\right\vert ^{\alpha^{\prime
}-\beta^{\prime}+\gamma^{\prime}}}\\
&  \leq\mathrm{const.}~\frac{1}{t^{\beta^{\prime}}}\frac{1}{1+\left(
\left\vert k\right\vert t^{1/2}\right)  ^{\alpha^{\prime}-\beta^{\prime
}+\gamma^{\prime}}}~,
\end{align*}
as claimed, and for $t>2$ we use that
\begin{align*}
&  \left(  1+\left(  \left\vert k\right\vert t^{1/2}\right)  ^{\alpha^{\prime
}-\beta^{\prime}+\gamma^{\prime}}\right)  e^{\mu\Lambda_{-}(t-1)}\left\vert
\Lambda_{-}t\right\vert ^{\beta^{\prime}}\left(  \dfrac{t-1}{t}\right)
^{\gamma^{\prime}}\\
&  \leq\mathrm{const.}\left(  1+\left(  \left\vert k\right\vert t^{1/2}%
\right)  ^{\alpha^{\prime}}\right)  e^{\frac{1}{2}\mu\Lambda_{-}t}\left\vert
\Lambda_{-}t\right\vert ^{\beta^{\prime}}\\
&  \leq\mathrm{const.}\left(  1+\frac{\left\vert k\right\vert ^{\alpha
^{\prime}}}{\left\vert \Lambda_{-}\right\vert ^{\alpha^{\prime}/2}}\left\vert
\Lambda_{-}t\right\vert ^{\alpha^{\prime}/2}\left\vert \Lambda_{-}t\right\vert
^{\beta^{\prime}}e^{\frac{1}{2}\mu\Lambda_{-}t}\right) \\
&  \leq\mathrm{const.}\left(  1+\frac{\left\vert k\right\vert ^{\alpha
^{\prime}}}{\left\vert \Lambda_{-}\right\vert ^{\alpha^{\prime}/2}}\right) \\
&  \leq\mathrm{const.}\left(  1+\left\vert k\right\vert ^{\alpha^{\prime}%
/2}\right)  \leq\mathrm{const.}\left(  1+\left\vert k\right\vert
^{\alpha^{\prime}}\right)  ~,
\end{align*}
and (\ref{decay1}) follows. We now prove (\ref{decay2}). For $1\leq t\leq2$
and $\left\vert k\right\vert \leq1$ we have that%
\begin{align*}
\frac{1}{1+\left\vert k\right\vert ^{\alpha^{\prime}}}e^{-\mu\left\vert
k\right\vert \left(  t-1\right)  }\left\vert k\right\vert ^{\beta^{\prime}%
}\left(  \dfrac{t-1}{t}\right)  ^{\gamma^{\prime}} &  \leq\mathrm{const.}\\
&  \leq\mathrm{const.}~\frac{1}{t^{\beta^{\prime}}}\frac{1}{1+\left(
\left\vert k\right\vert t\right)  ^{\alpha^{\prime}-\beta^{\prime}%
+\gamma^{\prime}}}~,
\end{align*}
and for $1\leq t\leq2$ and $\left\vert k\right\vert >1$ we have that%
\begin{align*}
\frac{1}{1+\left\vert k\right\vert ^{\alpha^{\prime}}}e^{-\mu\left\vert
k\right\vert \left(  t-1\right)  }\left\vert k\right\vert ^{\beta^{\prime}%
}\left(  \dfrac{t-1}{t}\right)  ^{\gamma^{\prime}} &  \leq\mathrm{const.}%
~\frac{1}{1+\left\vert k\right\vert ^{\alpha^{\prime}}}e^{-\mu\left\vert
k\right\vert \left(  t-1\right)  }\left(  \left\vert k\right\vert \left(
t-1\right)  \right)  ^{\gamma^{\prime}}~\left\vert k\right\vert ^{\beta
^{\prime}-\gamma^{\prime}}\\
&  \leq\mathrm{const.}~\frac{1}{1+\left\vert k\right\vert ^{\alpha^{\prime}}%
}~\left\vert k\right\vert ^{\beta^{\prime}-\gamma^{\prime}}\\
&  \leq\mathrm{const.}~\frac{1}{1+\left\vert k\right\vert ^{\alpha^{\prime
}-\beta^{\prime}+\gamma^{\prime}}}\\
&  \leq\mathrm{const.}~\frac{1}{t^{\beta^{\prime}}}\frac{1}{1+\left(
\left\vert k\right\vert t\right)  ^{\alpha^{\prime}-\beta^{\prime}%
+\gamma^{\prime}}}~.
\end{align*}
Finally, for $t>2$ we use that
\begin{align*}
&  \left(  1+\left(  \left\vert k\right\vert t\right)  ^{\alpha^{\prime}%
-\beta^{\prime}+\gamma^{\prime}}\right)  e^{-\mu\left\vert k\right\vert
\left(  t-1\right)  }\left(  \left\vert k\right\vert t\right)  ^{\beta
^{\prime}}\left(  \dfrac{t-1}{t}\right)  ^{\gamma^{\prime}}\\
&  \leq\mathrm{const.}\left(  1+\left(  \left\vert k\right\vert t\right)
^{\alpha^{\prime}-\beta^{\prime}+\gamma^{\prime}}\right)  e^{-\frac{1}{2}%
\mu\left\vert k\right\vert t}\left(  \left\vert k\right\vert t\right)
^{\beta^{\prime}}\\
&  \leq\mathrm{const.}\leq\mathrm{const.}\left(  1+\left\vert k\right\vert
^{\alpha^{\prime}}\right)  ~,
\end{align*}
and (\ref{decay2}) follows.
\end{proof}

\bigskip

We are now ready to prove (\ref{bpm1})-(\ref{bpm5}).

\subsection{Bound on $\omega_{0}$}

By definition (\ref{omega0}) of $\omega_{0}$ the inequality (\ref{bpm1})
follows from (\ref{bpm3}).

\subsection{Bound on $u_{0}$}

We write $u_{0}=\sum_{i=1}^{2}u_{0,i}$, with $u_{0,i}$ the $i$-th term in
(\ref{u0}), and we bound each of the terms individually. The inequality
(\ref{bpm3}) then follows using the triangle inequality.

\begin{proposition}
For all $\alpha\geq0$ we have the bounds%
\begin{align}
\left\vert u_{0,1}(k,t)\right\vert  & \leq\varepsilon\mu_{a+1}%
(k,t)~,\label{u0b1}\\
\left\vert u_{0,2}(k,t)\right\vert  & \leq\varepsilon\mu_{a+1}%
(k,t)~,\label{u0b2}%
\end{align}
uniformly in $t\geq1$ and $k\in\mathbf{R}$.
\end{proposition}

For $u_{0,1}$ we have%
\[
\left\vert -\omega_{-}^{\ast}(k)e^{\Lambda_{-}(t-1)}\right\vert \leq
\varepsilon\mu_{a+1}(k,1)e^{\Lambda_{-}(t-1)}~,
\]
and (\ref{u0b1}) follows using Proposition \ref{basic}. Next, splitting the
integral defining of $u_{0,2}$ in two parts we find that%
\begin{align*}
\left\vert \frac{1}{\Lambda_{0}}\int_{1}^{\frac{t+1}{2}}e^{\Lambda_{-}%
(t-s)}q_{1}(k,s)~ds\right\vert  & \leq\varepsilon\mu_{a+1}(k,1)e^{\Lambda
_{-}\frac{t-1}{2}}\int_{1}^{\frac{t+1}{2}}\frac{1}{s^{3/2}}ds\\
& \leq\varepsilon\mu_{a+1}(k,1)e^{\Lambda_{-}\frac{t-1}{2}}\left(  \frac
{t-1}{t}\right) \\
& \leq\varepsilon\mu_{a+1}(k,t)~,
\end{align*}
and that%
\begin{align*}
\left\vert \frac{1}{\Lambda_{0}}\int_{\frac{t+1}{2}}^{t}e^{\Lambda_{-}%
(t-s)}q_{1}(k,s)~ds\right\vert  & \leq\varepsilon\frac{1}{\Lambda_{0}}\mu
_{a}(k,t)\int_{\frac{t+1}{2}}^{t}\frac{1}{s^{3/2}}ds\\
& \leq\frac{\varepsilon}{t^{1/2}}\frac{1}{\Lambda_{0}}\mu_{a}(k,t)\\
& \leq\varepsilon\mu_{a+1}(k,t)~,
\end{align*}
which proves (\ref{u0b2}) using the triangle inequality.

\subsection{Bound on $\omega_{1}$}

We write $\omega_{1}=\sum_{i=1}^{3}\omega_{1,i}$, with $\omega_{1,i}$ the
$i$-th term in (\ref{omega1}), and we bound each of the terms individually.
The inequality (\ref{bpm2}) then follows using the triangle inequality.

\begin{proposition}
For all $\alpha\geq0$ we have the bounds%
\begin{align}
\left\vert \omega_{1,1}(k,t)\right\vert  & \leq\frac{\varepsilon}{t^{3/2}}%
\mu_{a}(k,t)~,\label{omega0b1}\\
\left\vert \omega_{1,2}(k,t)\right\vert  & \leq\frac{\varepsilon}{t}\mu
_{a}(k,t)~,\label{omega0b2}\\
\left\vert \omega_{1,3}(k,t)\right\vert  & \leq\frac{\varepsilon}{t}\mu
_{a+1}(k,t)~,\label{omega0b3}%
\end{align}
uniformly in $t\geq1$ and $k\in\mathbf{R}$.
\end{proposition}

For $\omega_{1,1}$ we have that%
\begin{align*}
\left\vert \frac{1}{\Lambda_{0}}\int_{t}^{\infty}e^{\Lambda_{+}(t-s)}%
ikq_{1}(k,s)~ds\right\vert  & \leq\frac{\varepsilon}{t^{3/2}}\frac{\left\vert
k\right\vert }{\Lambda_{0}}\mu_{a}(k,t)\int_{t}^{\infty}e^{\Lambda_{+}%
(t-s)}ds\\
& \leq\frac{\varepsilon}{t^{3/2}}\frac{1}{\Lambda_{+}}\frac{\left\vert
k\right\vert }{\Lambda_{0}}\mu_{a}(k,t)\\
& \leq\frac{\varepsilon}{t^{3/2}}\mu_{a}(k,t)~,
\end{align*}
which proves (\ref{omega0b1}). Similarly, to prove the bound on $\omega_{1,2}%
$, we use that by definition (\ref{q0}) of $q_{0}$%
\begin{align}
\left\vert q_{0}(k,t)\right\vert  & \leq\frac{\varepsilon}{t^{3/2}}\mu
_{a}(k,t)+\frac{\varepsilon}{t^{1/2}}\left\vert k\right\vert \mu
_{a+1}(k,t)\label{q0b1}\\
& \leq\frac{\varepsilon}{t}\mu_{a}(k,t)~,\label{q0b2}%
\end{align}
and therefore
\begin{align*}
\left\vert \frac{\Lambda_{+}}{\Lambda_{0}}\int_{t}^{\infty}e^{\Lambda
_{+}(t-s)}q_{0}(k,s)~ds\right\vert  & \leq\frac{\varepsilon}{t}\frac
{\Lambda_{+}}{\Lambda_{0}}\mu_{a}(k,t)\int_{t}^{\infty}e^{\Lambda_{+}%
(t-s)}ds\\
& \leq\frac{\varepsilon}{t}\frac{1}{\Lambda_{0}}\mu_{a}(k,t)~,
\end{align*}
which proves (\ref{omega0b2}). The integral defining $\omega_{1,3}$ we split
in two parts. Using (\ref{q0b1}) and the identity $\left\vert k\right\vert
=\left\vert \Lambda_{-}\right\vert ^{1/2}\left\vert \Lambda_{+}\right\vert
^{1/2}$, we find that
\begin{align*}
\left\vert \frac{\Lambda_{-}}{\Lambda_{0}}\int_{1}^{\frac{t+1}{2}}%
e^{\Lambda_{-}(t-s)}q_{0}(k,s)~ds\right\vert  & \leq\varepsilon\mu
_{a+1}(k,1)e^{\Lambda_{-}\frac{t-1}{2}}\left\vert \Lambda_{-}\right\vert
\int_{1}^{\frac{t+1}{2}}\frac{1}{s^{3/2}}ds\\
& +\varepsilon\mu_{a+2}(k,1)e^{\Lambda_{-}\frac{t-1}{2}}\left\vert \Lambda
_{-}\right\vert \left\vert k\right\vert \int_{1}^{\frac{t+1}{2}}\frac
{1}{s^{1/2}}ds\\
& \leq\varepsilon\mu_{a+1}(k,1)e^{\Lambda_{-}\frac{t-1}{2}}\left\vert
\Lambda_{-}\right\vert \left(  \frac{t-1}{t}\right) \\
& +\varepsilon t^{1/2}\mu_{a+3/2}(k,1)e^{\Lambda_{-}\frac{t-1}{2}}\left\vert
\Lambda_{-}\right\vert ^{3/2}\left(  \frac{t-1}{t}\right) \\
& \leq\frac{\varepsilon}{t}\mu_{a+1}(k,t)~,
\end{align*}
and, using (\ref{q0b2}), we find that%
\begin{align*}
\left\vert \frac{\Lambda_{-}}{\Lambda_{0}}\int_{\frac{t+1}{2}}^{t}%
e^{\Lambda_{-}(t-s)}q_{0}(k,s)~ds\right\vert  & \leq\frac{\varepsilon}{t}%
\frac{1}{\Lambda_{0}}\mu_{a}(k,t)\int_{\frac{t+1}{2}}^{t}e^{\Lambda_{-}%
(t-s)}\left\vert \Lambda_{-}\right\vert ds\\
& \leq\frac{\varepsilon}{t}\frac{1}{\Lambda_{0}}\mu_{a}(k,t)~,
\end{align*}
and (\ref{omega0b3}) follows using the triangle inequality.

\subsection{Bound on $u_{1}$}

We write $u_{1}=\sum_{i=1}^{10}u_{1,i}$, with $u_{1,i}$ the $i$-th term in
(\ref{u1}), and we bound each of the terms individually. The inequality
(\ref{bpm4}) then follows using the triangle inequality.

\begin{proposition}
\label{p11}For all $\alpha\geq0$ we have the bounds%
\begin{align}
\left\vert u_{1,1}(k,t)\right\vert  & \leq\varepsilon\bar{\mu}_{a+1}%
(k,t)~,\label{u01b1}\\
\left\vert u_{2,2}(k,t)\right\vert  & \leq\frac{\varepsilon}{t}\mu
_{a}(k,t)~,\label{u01b2}\\
\left\vert u_{1,3}(k,t)\right\vert  & \leq\frac{\varepsilon}{t}\mu
_{a+1}(k,t)~,\label{u01b3}\\
\left\vert u_{1,4}(k,t)\right\vert  & \leq\varepsilon\bar{\mu}_{a+1}%
(k,t)+\frac{\varepsilon}{t^{1/2}}\mu_{a}(k,t)~,\label{u01b4}\\
\left\vert u_{1,5}(k,t)\right\vert  & \leq\frac{\varepsilon}{t^{1/2}}\mu
_{a}(k,t)~,\label{u01b5}\\
\left\vert u_{1,6}(k,t)\right\vert  & \leq\frac{\varepsilon}{t}\mu
_{a+1}(k,t)~,\label{u01b6}\\
\left\vert u_{1,7}(k,t)\right\vert  & \leq\frac{\varepsilon}{t^{1/2}}%
\mu_{a+1/2}(k,t)~,\label{u01b7}\\
\left\vert u_{1,8}(k,t)\right\vert  & \leq\varepsilon\bar{\mu}_{a+1}%
(k,t)+\frac{\varepsilon}{t^{1/2}}\mu_{a}(k,t)~,\label{u01b8}\\
\left\vert u_{1,9}(k,t)\right\vert  & \leq\frac{\varepsilon}{t^{1/2}}\mu
_{a}(k,t)~,\label{u01b9}\\
\left\vert u_{1,10}(k,t)\right\vert  & \leq\frac{\varepsilon}{t}\mu
_{a}(k,t)~,\label{u01b10}%
\end{align}
uniformly in $t\geq1$ and $k\in\mathbf{R}$.
\end{proposition}

For $u_{1,1}$ we have that%
\[
\left\vert \left(  u_{-,1}^{\ast}(k)-i\sigma(k)u_{-,2}^{\ast}(k)\right)
e^{-\left\vert k\right\vert (t-1)}\right\vert \leq\varepsilon\mu_{\alpha
+1}(k,1)e^{-\left\vert k\right\vert (t-1)}~,
\]
and (\ref{u01b1}) follows using Proposition \ref{basic}. Next, for $u_{1,2}$
we have that
\[
\left\vert \omega_{-}^{\ast}(k)\Lambda_{-}e^{\Lambda_{-}(t-1)}\right\vert
\leq\varepsilon\mu_{\alpha+1}(k,1)\left\vert \Lambda_{-}\right\vert
e^{\Lambda_{-}(t-1)}~,
\]
and (\ref{u01b2}) follows using Proposition \ref{basic}. The integral defining
$u_{1,3}$ we split in two parts. We have that%
\begin{align*}
\left\vert \frac{\Lambda_{-}}{\Lambda_{0}}\int_{1}^{\frac{t+1}{2}}%
e^{\Lambda_{-}(t-s)}q_{1}(k,s)~ds\right\vert  & \leq\varepsilon\mu_{\alpha
+1}(k,1)\left\vert \Lambda_{-}\right\vert e^{\Lambda_{-}\frac{t-1}{2}}\int
_{1}^{\frac{t+1}{2}}\frac{1}{s^{3/2}}ds\\
& \leq\varepsilon\mu_{\alpha+1}(k,1)e^{\Lambda_{-}\frac{t-1}{2}}\left\vert
\Lambda_{-}\right\vert \left(  \frac{t-1}{t}\right) \\
& \leq\frac{\varepsilon}{t}\mu_{\alpha+1}(k,t)~,
\end{align*}
and that
\begin{align*}
\left\vert \frac{\Lambda_{-}}{\Lambda_{0}}\int_{\frac{t+1}{2}}^{t}%
e^{\Lambda_{-}(t-s)}q_{1}(k,s)~ds\right\vert  & \leq\frac{\varepsilon}%
{t^{3/2}}\frac{1}{\Lambda_{0}}\mu_{\alpha}(k,t)\int_{\frac{t+1}{2}}%
^{t}e^{\Lambda_{-}(t-s)}\left\vert \Lambda_{-}\right\vert ~ds\\
& \leq\frac{\varepsilon}{t^{3/2}}\frac{1}{\Lambda_{0}}\mu_{\alpha}(k,t)\\
& \leq\frac{\varepsilon}{t}\mu_{\alpha+1}(k,t)~,
\end{align*}
and (\ref{u01b3}) follows. The integral defining $u_{1,4}$ we also split in
two parts. We have that
\begin{align*}
\left\vert \frac{1}{2}\int_{1}^{\frac{t+1}{2}}e^{-\left\vert k\right\vert
(t-s)}q_{1}(k,s)~ds\right\vert  & \leq\varepsilon e^{-\left\vert k\right\vert
\frac{t-1}{2}}\mu_{\alpha}(k,1)\int_{1}^{\frac{t+1}{2}}\frac{1}{s^{3/2}}~ds\\
& \leq\varepsilon\mu_{\alpha}(k,1)e^{-\left\vert k\right\vert \frac{t-1}{2}%
}\left(  \frac{t-1}{t}\right) \\
& \leq\varepsilon\bar{\mu}_{\alpha+1}(k,t)~,
\end{align*}
and that
\begin{align*}
\left\vert \frac{1}{2}\int_{\frac{t+1}{2}}^{t}e^{-\left\vert k\right\vert
(t-s)}q_{1}(k,s)~ds\right\vert  & \leq\varepsilon\mu_{\alpha}(k,t)\int
_{\frac{t+1}{2}}^{t}\frac{1}{s^{3/2}}~ds\\
& \leq\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}(k,t)~,
\end{align*}
and (\ref{u01b4}) follows. Next, to bound $u_{1,5}$, we use that
\begin{align*}
\left\vert \frac{1}{2}\int_{t}^{\infty}e^{\left\vert k\right\vert (t-s)}%
q_{1}(k,s)~ds\right\vert  & \leq\varepsilon\mu_{\alpha}(k,t)\int_{t}^{\infty
}\frac{1}{s^{3/2}}~ds\\
& \leq\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}(k,t)~,
\end{align*}
and (\ref{u01b5}) follows. Next, for $u_{1,6}$ we have that
\begin{align*}
\left\vert \frac{\Lambda_{-}}{\Lambda_{0}}\int_{t}^{\infty}e^{\Lambda
_{+}(t-s)}q_{1}(k,s)~ds\right\vert  & \leq\frac{\varepsilon}{t^{3/2}}%
\frac{\left\vert \Lambda_{-}\right\vert }{\Lambda_{0}}\mu_{\alpha}%
(k,t)\int_{t}^{\infty}e^{\Lambda_{+}(t-s)}~ds\\
& \leq\frac{\varepsilon}{t^{3/2}}\frac{\left\vert \Lambda_{-}\right\vert
}{\Lambda_{0}\Lambda_{+}}\mu_{\alpha}(k,t)\\
& \leq\frac{\varepsilon}{t}\mu_{\alpha+1}(k,t)~,
\end{align*}
and (\ref{u01b6}) follows. To bound $u_{1,7}$ we split the integral into two
parts and estimate the contributions from $q_{0,0}$ and $q_{0,1}$ separately.
For the contribution to $u_{1,7}$ coming from $q_{0,0}$ we have that%
\begin{align*}
\left\vert \frac{-ik}{\Lambda_{0}}\int_{1}^{\frac{t+1}{2}}e^{\Lambda_{-}%
(t-s)}q_{0,0}(k,s)~ds\right\vert  & \leq\frac{\left\vert k\right\vert
}{\Lambda_{0}}\int_{1}^{\frac{t+1}{2}}e^{\Lambda_{-}(t-s)}\frac{\varepsilon
}{s^{3/2}}\mu_{a}(k,s)~ds\\
& \leq\varepsilon\frac{\left\vert k\right\vert }{\Lambda_{0}}e^{\Lambda
_{-}\frac{t-1}{2}}\mu_{a}(k,1)\int_{1}^{\frac{t+1}{2}}\frac{1}{s^{3/2}}~ds\\
& \leq\frac{\varepsilon}{t^{1/2}}\mu_{a+1/2}(k,t)~,
\end{align*}
and that
\begin{align*}
\left\vert \frac{-ik}{\Lambda_{0}}\int_{\frac{t+1}{2}}^{t}e^{\Lambda_{-}%
(t-s)}q_{0,0}(k,s)~ds\right\vert  & \leq\frac{\left\vert k\right\vert
}{\Lambda_{0}}\int_{\frac{t+1}{2}}^{t}e^{\Lambda_{-}(t-s)}\frac{\varepsilon
}{s^{3/2}}\mu_{a}(k,s)~ds\\
& \leq\frac{\varepsilon}{t^{3/2}}\mu_{a}(k,t)\frac{\Lambda_{+}^{1/2}}%
{\Lambda_{0}}\int_{\frac{t+1}{2}}^{t}e^{\Lambda_{-}(t-s)}\left\vert
\Lambda_{-}\right\vert ^{1/2}~ds\\
& \leq\frac{\varepsilon}{t^{3/2}}\frac{1}{\Lambda_{0}^{1/2}}\mu_{a}%
(k,t)\int_{\frac{t+1}{2}}^{t}\frac{1}{\sqrt{t-s}}~ds\\
& \leq\frac{\varepsilon}{t}\frac{1}{\Lambda_{0}^{1/2}}\mu_{a}(k,t)~,
\end{align*}
and for the contribution to $u_{1,7}$ coming from $q_{0,1}$ we have that
\begin{align*}
\left\vert \frac{-ik}{\Lambda_{0}}\int_{1}^{\frac{t+1}{2}}e^{\Lambda_{-}%
(t-s)}q_{0,1}(k,s)~ds\right\vert  & \leq\frac{\left\vert k\right\vert
}{\Lambda_{0}}\int_{1}^{\frac{t+1}{2}}e^{\Lambda_{-}(t-s)}\frac{\varepsilon
}{s^{1/2}}\left\vert k\right\vert \mu_{a+1}(k,s)~ds\\
& \leq\varepsilon\mu_{a+1}(k,1)e^{\Lambda_{-}\frac{t-1}{2}}\left\vert
\Lambda_{-}\right\vert \int_{1}^{\frac{t+1}{2}}\frac{\varepsilon}{s^{1/2}%
}~ds\\
& \leq\frac{\varepsilon}{t^{1/2}}\mu_{a+1}(k,1)e^{\Lambda_{-}\frac{t-1}{2}%
}\left\vert \Lambda_{-}\right\vert (t-1)\\
& \leq\frac{\varepsilon}{t^{1/2}}\mu_{a+1}(k,t)~,
\end{align*}
and that
\begin{align*}
\left\vert \frac{-ik}{\Lambda_{0}}\int_{\frac{t+1}{2}}^{t}e^{\Lambda_{-}%
(t-s)}q_{0,1}(k,s)~ds\right\vert  & \leq\frac{\left\vert k\right\vert
}{\Lambda_{0}}\int_{\frac{t+1}{2}}^{t}e^{\Lambda_{-}(t-s)}\frac{\varepsilon
}{s^{1/2}}\left\vert k\right\vert \mu_{a+1}(k,s)~ds\\
& \leq\frac{\varepsilon}{t^{1/2}}\mu_{a+1}(k,t)\int_{\frac{t+1}{2}}%
^{t}e^{\Lambda_{-}(t-s)}\left\vert \Lambda_{-}\right\vert ~ds\\
& \leq\frac{\varepsilon}{t^{1/2}}\mu_{a+1}(k,t)~,
\end{align*}
and (\ref{u01b7}) follows using the triangle inequality. The proof of the
estimate for the $q_{0,0}$-contribution to $u_{1,8}$ is identical to the proof
of (\ref{u01b4}), and we therefore have that%
\[
\left\vert \frac{1}{2}i\sigma(k)\int_{1}^{t}e^{-\left\vert k\right\vert
(t-s)}q_{0,0}(k,s)~ds\right\vert \leq\varepsilon\bar{\mu}_{\alpha
+1}(k,t)+\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}(k,t)~.
\]
To estimate the contribution of $q_{0,1}$ to $u_{1,8}$ we split the integral
into two and we get%
\begin{align*}
\left\vert \frac{1}{2}i\sigma(k)\int_{1}^{\frac{t+1}{2}}e^{-\left\vert
k\right\vert (t-s)}q_{0,1}(k,s)~ds\right\vert  & \leq\int_{1}^{\frac{t+1}{2}%
}e^{-\left\vert k\right\vert (t-s)}\frac{\varepsilon}{s^{1/2}}\left\vert
k\right\vert \mu_{a+1}(k,s)~ds\\
& \leq\varepsilon\mu_{a+1}(k,1)\left\vert k\right\vert e^{-\left\vert
k\right\vert \frac{t-1}{2}}\int_{1}^{\frac{t+1}{2}}\frac{1}{s^{1/2}}ds\\
& \leq\varepsilon t^{1/2}\mu_{a+1}(k,1)e^{-\left\vert k\right\vert \frac
{t-1}{2}}\left\vert k\right\vert \left(  \frac{t-1}{t}\right) \\
& \leq\frac{\varepsilon}{t^{1/2}}\bar{\mu}_{a+1}(k,t)~,
\end{align*}
and%
\begin{align*}
\left\vert \frac{1}{2}i\sigma(k)\int_{\frac{t+1}{2}}^{t}e^{-\left\vert
k\right\vert (t-s)}q_{0,1}(k,s)~ds\right\vert  & \leq\int_{\frac{t+1}{2}}%
^{t}e^{-\left\vert k\right\vert (t-s)}\frac{\varepsilon}{s^{1/2}}\left\vert
k\right\vert \mu_{a+1}(k,s)~ds\\
& \leq\frac{\varepsilon}{t^{1/2}}\mu_{a+1}(k,t)\int_{\frac{t+1}{2}}%
^{t}e^{-\left\vert k\right\vert (t-s)}\left\vert k\right\vert ~ds\\
& \leq\frac{\varepsilon}{t^{1/2}}\mu_{a+1}(k,t)~,
\end{align*}
and (\ref{u01b8}) follows using the triangle inequality. The proof of the
estimate for the $q_{0,0}$-contribution to $u_{1,9}$ is identical to the proof
of (\ref{u01b5}), so that%
\[
\left\vert \frac{1}{2}i\sigma(k)\int_{t}^{\infty}e^{\left\vert k\right\vert
(t-s)}q_{0,0}(k,s)~ds\right\vert \leq\frac{\varepsilon}{t^{1/2}}\mu_{\alpha
}(k,t)~.
\]
For the contribution of $q_{0,1}$ to $u_{1,9}$ we get (this is a more
elementary but less precise bound than (\ref{ccbound}))%
\begin{align*}
\left\vert \frac{1}{2}i\sigma(k)\int_{t}^{\infty}e^{\left\vert k\right\vert
(t-s)}\left(  -ikq_{0,1}(k,s)\right)  ~ds\right\vert  & \leq\int_{t}^{\infty
}e^{\left\vert k\right\vert (t-s)}\frac{\varepsilon}{s^{1/2}}\left\vert
k\right\vert \mu_{a+1}(k,s)~ds\\
& \leq\frac{\varepsilon}{t^{1/2}}\mu_{a+1}(k,t)\int_{t}^{\infty}e^{\left\vert
k\right\vert (t-s)}\left\vert k\right\vert ~ds\\
& \leq\frac{\varepsilon}{t^{1/2}}\mu_{a+1}(k,t)~,
\end{align*}
and (\ref{u01b9}) follows using the triangle inequality. Finally, since
$u_{1,10}=-ik/\Lambda_{0}~\omega_{1,2}$, and since $\left\vert -ik/\Lambda
_{0}\right\vert \leq\mathrm{const.}$, the proof of the estimate (\ref{u01b10})
is identical to the proof of (\ref{omega0b2})\textrm{.}

\subsection{Bound on $v$}

We write $v=\sum_{i=1}^{6}v_{i}$, with $v_{i}$ the $i$-th term in (\ref{v_t}),
and we bound each of the terms individually. The inequality (\ref{bpm5}) then
follows using the triangle inequality.

\begin{proposition}
For all $\alpha\geq0$ we have the bounds%
\begin{align}
\left\vert v_{1}(k,t)\right\vert  & \leq\frac{\varepsilon}{t^{1/2}}\mu
_{a}(k,t)~,\label{vb1}\\
\left\vert v_{2}(k,t)\right\vert  & \leq\varepsilon\bar{\mu}_{\alpha
+1}(k,t)~,\label{vb2}\\
\left\vert v_{3}(k,t)\right\vert  & \leq\varepsilon\bar{\mu}_{\alpha
+1}(k,t)+\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}(k,t)~,\label{vb3}\\
\left\vert v_{4}(k,t)\right\vert  & \leq\frac{\varepsilon}{t^{1/2}}\mu
_{a}(k,t)~,\label{vb4}\\
\left\vert v_{5}(k,t)\right\vert  & \leq\varepsilon\bar{\mu}_{a+1}%
(k,t)+\frac{\varepsilon}{t^{1/2}}\mu_{a}(k,t)~,\label{vb5}\\
\left\vert v_{6}(k,t)\right\vert  & \leq\frac{\varepsilon}{t^{1/2}}\mu
_{a}(k,t)~,\label{vb6}%
\end{align}
uniformly in $k\in\mathbf{R}$ and $t\geq1$.
\end{proposition}

Inequality (\ref{vb1}) follows from (\ref{bpm1}) and (\ref{bpm2}) using the
triangle inequality. Next, after substitution of (\ref{umt}) and (\ref{um})
into $v_{2}$ we find that $v_{2}(k,t)=i\sigma(k)u_{1,1}(k,t).$ Therefore,
\[
\left\vert v_{2}(k,t)\right\vert \leq\left\vert u_{1,1}(k,t)\right\vert ~,
\]
and (\ref{vb2}) therefore follows from (\ref{u01b1}). Similarly, (\ref{vb3}),
(\ref{vb4}), (\ref{vb5}) and (\ref{vb6}) follow from (\ref{u01b4}),
(\ref{u01b5}), (\ref{u01b8}) and (\ref{u01b9}), since
\begin{align*}
v_{3}(k,t)  & =i\sigma(k)~u_{1,4}(k,t)~,\\
v_{4}(k,t)  & =-i\sigma(k)~u_{1,5}(k,t)~,\\
v_{5}(k,t)  & =-i\sigma(k)~u_{1,8}(k,t)~,\\
v_{6}(k,t)  & =i\sigma(k)~u_{1,9}(k,t)~.
\end{align*}
This completes the proof of Proposition \ref{pp11}.

\section{Appendix II}

In this appendix we give a proof of Proposition \ref{pp12}.

\subsection{Bounds on convolutions}

\begin{proposition}
\label{c1}Let $\alpha$, $\beta>1$, $p\geq q\geq0$ and let $a$ be a piecewise
continuous, and $b$ be a continuous function from $\mathbf{R}\times
\lbrack1,\infty)$ to $\mathbf{C}$ satisfying the bounds (see (\ref{mu}) for
the definition of $\mu_{\alpha}^{p}$ and $\mu_{\beta}^{q}$, respectively),%
\begin{align*}
\left\vert a(k,t)\right\vert  & \leq\mu_{\alpha}^{p}(k,t)~,\\
\left\vert b(k,t)\right\vert  & \leq\mu_{\beta}^{q}(k,t)~.
\end{align*}
Then, the convolution $a\ast b$ is a continuous function from $\mathbf{R}%
\times\lbrack1,\infty)$ to $\mathbf{C}$ and we have the bounds%
\begin{align}
\left\vert \left(  a\ast b\right)  (k,t)\right\vert  & \leq\mathrm{const.}%
\left(  \frac{1}{t^{p}}\mu_{\beta}^{q}(k,t)+\left\vert k\right\vert
\mu_{\alpha}^{p}(k,t)\right)  ~,\label{c1b1}\\
\left\vert \left(  a\ast b\right)  (k,t)\right\vert  & \leq\mathrm{const.}%
~\frac{1}{t^{p}}\mu_{\min\{\alpha-1,\beta\}}^{q}(k,t)~,\label{c1b2}\\
\left\vert \left(  a\ast b\right)  (k,t)\right\vert  & \leq\mathrm{const.}%
\left(  \frac{1}{t^{p}}\mu_{\beta}^{q}(k,t)+\frac{1}{t^{q}}\mu_{\alpha}%
^{p}(k,t)\right)  ~,\label{c1b3}\\
\left\vert \left(  a\ast b\right)  (k,t)\right\vert  & \leq\mathrm{const.}%
~\frac{1}{t^{\min\{p,q\}}}\mu_{\min\{\alpha,\beta\}}^{\min\{p,q\}}%
(k,t)~,\label{c1b4}%
\end{align}
uniformly in $t\geq1$, $k\in\mathbf{R}$.
\end{proposition}

Continuity is elementary. The bound (\ref{c1b2}) follows immediately from
(\ref{c1b1}) using that $\left\vert k\right\vert \mu_{\alpha}^{p}%
(k,t)\leq\mathrm{const.}~\mu_{\alpha-1}^{p}(k,t)/t^{p}$, and (\ref{c1b4})
follows from (\ref{c1b3}). We now prove (\ref{c1b1}). Let $k\geq0$. Then, we
have for $a\ast b$%
\begin{align}
\left\vert \left(  a\ast b\right)  (k,t)\right\vert  & \leq\int_{-\infty
}^{\infty}\mu_{\alpha}^{p}(k^{\prime},t)\mu_{\beta}^{q}(k-k^{\prime
},t)~dk^{\prime}\nonumber\\
& \leq\mu_{\beta}^{q}(k/2,t)\int_{-\infty}^{k/2}\mu_{\alpha}^{p}(k^{\prime
},t)~dk^{\prime}\nonumber\\
& +\int_{k/2}^{3k/2}\mu_{\alpha}^{p}(k^{\prime},t)~dk^{\prime}+\mu_{\beta}%
^{q}(k/2,t)\int_{3k/2}^{\infty}\mu_{\alpha}^{p}(k^{\prime},t)~dk^{\prime
}\nonumber\\
& \leq\mathrm{const.}\left(  \frac{1}{t^{p}}\mu_{\beta}^{q}(k,t)+\left\vert
k\right\vert \mu_{\alpha}^{p}(k/2,t)\right)  ~,\label{comp1}%
\end{align}
and (\ref{c1b1}) follows for $k\geq0$. Similarly we have for $k<0$,
\begin{align*}
\left\vert \left(  a\ast b\right)  (k,t)\right\vert  & \leq\int_{-\infty
}^{\infty}\mu_{\alpha}^{p}(k^{\prime},t)\mu_{\beta}^{q}(k-k^{\prime
},t)~dk^{\prime}\\
& \leq\mu_{\beta}^{q}(k/2,t)\int_{-\infty}^{3k/2}\mu_{\alpha}^{p}(k^{\prime
},t)~dk^{\prime}\\
& +\int_{3k/2}^{k/2}\mu_{\alpha}^{p}(k^{\prime},t)~dk^{\prime}+\mu_{\beta}%
^{q}(k/2,t)\int_{k/2}^{\infty}\mu_{\alpha}^{p}(k^{\prime},t)~dk^{\prime}\\
& \leq\mathrm{const.}\left(  \frac{1}{t^{p}}\mu_{\beta}^{q}(k,t)+\left\vert
k\right\vert \mu_{\alpha}^{p}(k/2,t)\right)  ~,
\end{align*}
and (\ref{c1b1}) follows for $k<0$. We now prove (\ref{c1b3}). Let $k\geq0$.
Then we have, cutting the integral into three parts as in (\ref{comp1}),%
\begin{align*}
\left\vert \left(  a\ast b\right)  (k,t)\right\vert  & \leq\mu_{\beta}%
^{q}(k/2,t)\int_{-\infty}^{k/2}\mu_{\alpha}^{p}(k^{\prime},t)~dk^{\prime}\\
& +\mu_{\alpha}^{p}(k/2,t)\int_{k/2}^{3k/2}\mu_{\beta}^{q}(k-k^{\prime
},t)~dk^{\prime}+\mu_{\beta}^{q}(k/2,t)\int_{3k/2}^{\infty}\mu_{\alpha}%
^{p}(k^{\prime},t)~dk^{\prime}\\
& \leq\mathrm{const.}\left(  \frac{1}{t^{p}}\mu_{\beta}^{q}(k,t)+\frac
{1}{t^{q}}\mu_{\alpha}^{p}(k,t)\right)  ~,
\end{align*}
and (\ref{c1b3}) follows for $k\geq0$. Similarly we have for $k<0$,
\begin{align*}
\left\vert \left(  a\ast b\right)  (k,t)\right\vert  & \leq\mu_{\beta}%
^{q}(k/2,t)\int_{-\infty}^{3k/2}\mu_{\alpha}^{p}(k^{\prime},t)~dk^{\prime}\\
& +\mu_{\alpha}^{p}(k/2,t)\int_{3k/2}^{k/2}\mu_{\beta}^{q}(k-k^{\prime
},t)~dk^{\prime}+\mu_{\beta}^{q}(k/2,t)\int_{k/2}^{\infty}\mu_{\alpha}%
^{p}(k^{\prime},t)~dk^{\prime}\\
& \leq\mathrm{const.}\left(  \frac{1}{t^{p}}\mu_{\beta}^{q}(k,t)+\frac
{1}{t^{q}}\mu_{\alpha}^{p}(k,t)\right)  ~,
\end{align*}
and (\ref{c1b3}) follows for $k<0$. This completes the proof of Proposition
\ref{c1}.

\subsection{Proof of Proposition \ref{pp12}}

We first prove the bound on $q_{0,0}$. Namely, using Proposition \ref{c1}, we
find from (\ref{bpm1}), (\ref{bpm2}), (\ref{bpm3}) and (\ref{bpm4}) that%
\begin{align*}
\left\vert \left(  u_{0}\ast\omega_{1}\right)  (k,t)\right\vert  & \leq
\frac{\varepsilon^{2}}{t^{3/2}}\mu_{a}(k,t)~,\\
\left\vert \left(  u_{1}\ast\omega_{0}\right)  (k,t)\right\vert  & \leq
\frac{\varepsilon^{2}}{t^{3/2}}\mu_{a}(k,t)~,\\
\left\vert \left(  u_{1}\ast\omega_{1}\right)  (k,t)\right\vert  & \leq
\frac{\varepsilon^{2}}{t^{2}}\mu_{a}(k,t)~,
\end{align*}
and (\ref{nqb1}) follows using the triangle inequality. Next we prove the
bound on $q_{0,1}$. Namely, using (\ref{c1b4}) we find from (\ref{bpm3}) that
\[
\left\vert \left(  u_{0}\ast u_{0}\right)  (k,t)\right\vert \leq
\frac{\varepsilon^{2}}{t^{1/2}}\mu_{a+1}(k,t)~,
\]
which proves (\ref{nqb2}). Finally, we prove the bound on $q_{1}$. First, the
bounds (\ref{bpm1}) and (\ref{bpm2}) imply using the triangle inequality,
that
\[
\left\vert \omega(k,t)\right\vert \leq\frac{\varepsilon}{t^{1/2}}\mu
_{a}(k,t)~,
\]
which, together with (\ref{bpm5}) and using Proposition \ref{c1} implies that
\[
\left\vert \left(  v\ast\omega\right)  (k,t)\right\vert \leq\frac
{\varepsilon^{2}}{t^{3/2}}\mu_{\alpha}(k,t)~,
\]
which proves (\ref{nqb3}). This completes the proof of Proposition \ref{pp12}.

\section{Appendix III}

In this appendix we prove Proposition \ref{prop12}. Let $\rho^{1}\equiv
(\rho_{1}^{1},\rho_{2}^{1},\rho_{3}^{1})$, $\rho^{2}\equiv(\rho_{1}^{2}%
,\rho_{2}^{2},\rho_{3}^{2})\in B_{\alpha}(\varepsilon_{0})$. Then, by
Proposition \ref{pp11} and Proposition \ref{pp12}, $\rho\equiv\mathcal{N}%
(\rho^{1})-\mathcal{N}(\rho^{2})$ is well defined and $\rho\in\mathcal{B}%
_{\alpha}$. Let $\rho\equiv(\rho_{1},\rho_{2},\rho_{3})$, and let $\omega
_{0}^{i}$, $\omega_{1}^{i}$, $\omega^{i}$, $u_{0}^{i}$, $u_{0}^{i}$,
$v_{0}^{i}$, $i=1,2$, be the quantities (\ref{omega0}), (\ref{omega1}),
(\ref{omega}), (\ref{u0}), (\ref{u1}) and (\ref{v_t}), computed from $\rho
^{1}$ and $\rho^{2}$, respectively. Using the identity $ab-\tilde{a}\tilde
{b}=(a-\tilde{a})b+\tilde{a}(b-\tilde{b})$ (distributive law) we find that%
\begin{align}
\rho_{1}  & =\frac{1}{2\pi}\left(  u_{0}^{1}\ast\omega_{1}^{1}+u_{1}^{1}%
\ast\omega_{0}^{1}+u_{1}^{1}\ast\omega_{1}^{1}\right) \nonumber\\
& -\frac{1}{2\pi}\left(  u_{0}^{2}\ast\omega_{1}^{2}+u_{1}^{2}\ast\omega
_{0}^{2}+u_{1}^{2}\ast\omega_{1}^{2}\right) \nonumber\\
& =\frac{1}{2\pi}\left[  \left(  u_{0}^{1}-u_{0}^{2}\right)  \ast\omega
_{1}^{1}+u_{0}^{2}\ast\left(  \omega_{1}^{1}-\omega_{1}^{2}\right)  \right]
\nonumber\\
& +\frac{1}{2\pi}\left[  \left(  u_{1}^{1}-u_{0}^{2}\right)  \ast\omega
_{0}^{1}+u_{1}^{2}\ast\left(  \omega_{0}^{1}-\omega_{0}^{2}\right)  \right]
\nonumber\\
& +\frac{1}{2\pi}\left[  \left(  u_{1}^{1}-u_{0}^{2}\right)  \ast\omega
_{1}^{1}+u_{1}^{2}\ast\left(  \omega_{1}^{1}-\omega_{1}^{2}\right)  \right]
~,\nonumber
\end{align}
and similarly that%
\begin{align*}
\rho_{2}  & =-\frac{1}{4\pi}\left[  \left(  u_{0}^{1}-u_{0}^{2}\right)  \ast
u_{0}^{1}+u_{0}^{2}\ast\left(  u_{0}^{1}-u_{0}^{2}\right)  \right]  ~,\\
\rho_{3}  & =\frac{1}{2\pi}\left[  \left(  v^{1}-v^{2}\right)  \ast\omega
^{1}+v^{2}\ast\left(  \omega^{1}-\omega^{2}\right)  \right]  ~.
\end{align*}
Therefore, and since the quantities $\omega_{0}^{i}$, $\omega_{1}^{i}$,
$\omega^{i}$, $u_{0}^{i}$, $u_{0}^{i}$, $v_{0}^{i}$, $i=1,2$ are linear
(respectively affine) in $\rho^{1}$ and $\rho^{2}$, the bound (\ref{lipschitz}%
) now follows \textit{mutatis mutandis} as in the proof of Proposition
\ref{pp11} and Proposition \ref{pp12}.

\section{Appendix IV}

In this appendix we prove (\ref{uas}) and (\ref{vas}).

\subsection{Asymptotic behavior of $u$}

Let%
\[
U(k,t)=\left(  -\omega_{-}^{\ast}(k)+\frac{1}{\Lambda_{0}}\int_{1}^{t}%
q_{1}(k,s)~ds\right)  e^{\Lambda_{-}(t-1)}~.
\]
Using the triangle inequality we get that%
\begin{align}
\left\vert u(k,t)-u_{as}(k,t)\right\vert  & \leq\left\vert U(k,t)-c~e^{-k^{2}%
t}\right\vert \nonumber\\
& +\left\vert u_{as}(k,t)-c~e^{-k^{2}t}\right\vert \nonumber\\
& +\left\vert u_{0}(k,t)-U(k,t)\right\vert \nonumber\\
& +\left\vert u_{1}(k,t)\right\vert ~.\label{ut}%
\end{align}
We bound each term in (\ref{ut}) separately. First, we have that%
\begin{equation}
\lim_{t\rightarrow\infty}U(\frac{k}{t^{1/2}},t)=c~e^{-k^{2}}~,\label{ld1}%
\end{equation}
and furthermore that%
\begin{align*}
\left\vert U(k,t)\right\vert  & \leq\left(  \varepsilon\mu_{\alpha
+1}(k,1)+\varepsilon\mu_{\alpha+1}(k,1)\int_{1}^{\infty}\frac{1}{s^{3/2}%
}ds\right)  e^{\Lambda_{-}(t-1)}\\
& \leq\varepsilon\mu_{\alpha+1}(k,t)~,
\end{align*}
so that%
\begin{equation}
\left\vert U(\frac{k}{t^{1/2}},t)\right\vert \leq\varepsilon\mu_{\alpha
+1}(k,1)~.\label{ld2}%
\end{equation}
From (\ref{ld1}) and (\ref{ld2}) it follows by the Lebesgue dominated
convergence theorem that%
\begin{align*}
& \lim_{t\rightarrow\infty}t^{1/2}\int_{\mathbf{R}}\left\vert
U(k,t)-c~e^{-k^{2}t}\right\vert ~dk\\
& =\lim_{t\rightarrow\infty}\int_{\mathbf{R}}\left\vert U(\frac{k}{t^{1/2}%
},t)-c~e^{-k^{2}}\right\vert ~dk\\
& =0~,
\end{align*}
as required. Next%
\begin{align*}
\lim_{t\rightarrow\infty}t^{1/2}\int_{\mathbf{R}}\left\vert u_{as}%
(k,t)-c~e^{-k^{2}t}\right\vert ~dk~  & \leq\lim_{t\rightarrow\infty}%
t^{1/2}\int_{\mathbf{R}}\left(  \left\vert d\right\vert +\left\vert
b\right\vert \right)  e^{-\left\vert k\right\vert t}~dk\\
& =\lim_{t\rightarrow\infty}t^{1/2}2\frac{\left\vert d\right\vert +\left\vert
b\right\vert }{t}=0~,
\end{align*}
as required. Next, using that $1-e^{x}\leq-x$ for all $x\leq0$, we find that%
\begin{align*}
\left\vert u_{0}(k,t)-U(k,t)\right\vert  & \leq\frac{1}{\Lambda_{0}}\int
_{1}^{t}\left(  e^{\Lambda_{-}(t-s)}-e^{\Lambda_{-}(t-1)}\right)  \left\vert
q_{1}(k,s)\right\vert ~ds\\
& \leq\frac{1}{\Lambda_{0}}\int_{1}^{t}e^{\Lambda_{-}(t-s)}\left(
1-e^{\Lambda_{-}(s-1)}\right)  \frac{\varepsilon}{s^{3/2}}\mu_{\alpha
}(k,s)~ds\\
& \leq-\frac{1}{\Lambda_{0}}\int_{1}^{t}e^{\Lambda_{-}(t-s)}\Lambda
_{-}(s-1)\frac{\varepsilon}{s^{3/2}}\mu_{\alpha}(k,s)~ds~,
\end{align*}
and therefore, since%
\begin{align*}
-\frac{1}{\Lambda_{0}}\int_{1}^{\frac{t+1}{2}}e^{\Lambda_{-}(t-s)}\Lambda
_{-}(s-1)\frac{\varepsilon}{s^{3/2}}\mu_{\alpha}(k,s)~ds  & \leq\varepsilon
\mu_{\alpha+1}(k,1)e^{\Lambda_{-}\frac{t-1}{2}}\Lambda_{-}\left(  \frac
{t-1}{t}\right)  ^{2}t^{1/2}\\
& \leq\frac{\varepsilon}{t^{1/2}}\mu_{\alpha+1}(k,t)~,
\end{align*}
and since
\begin{align*}
-\frac{1}{\Lambda_{0}}\int_{\frac{t+1}{2}}^{t}e^{\Lambda_{-}(t-s)}\Lambda
_{-}(s-1)\frac{\varepsilon}{s^{3/2}}\mu_{\alpha}(k,s)~ds  & \leq
-\frac{\varepsilon}{t^{1/2}}\frac{1}{\Lambda_{0}}\mu_{\alpha}(k,t)\int
_{\frac{t+1}{2}}^{t}e^{\Lambda_{-}(t-s)}\Lambda_{-}~ds\\
& \leq\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}(k,t)~,
\end{align*}
it follows that%
\begin{align*}
\lim_{t\rightarrow\infty}t^{1/2}\int_{\mathbf{R}}\left\vert u_{0}%
(k,t)-U(k,t)\right\vert ~dk  & \leq\lim_{t\rightarrow\infty}t^{1/2}%
\int_{\mathbf{R}}\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}(k,t)~dk\\
& \leq\lim_{t\rightarrow\infty}t^{1/2}\frac{\varepsilon}{t}=0~,
\end{align*}
as required. Finally, we have that%
\begin{align*}
\lim_{t\rightarrow\infty}t^{1/2}\int_{\mathbf{R}}\left\vert u_{1}%
(k,t)\right\vert ~dk  & \leq\lim_{t\rightarrow\infty}t^{1/2}\int_{\mathbf{R}%
}\left(  \varepsilon\bar{\mu}_{\alpha}(k,t)+\frac{\varepsilon}{t^{1/2}}%
\mu_{\alpha}(k,t)\right)  ~dk\\
& \leq\lim_{t\rightarrow\infty}t^{1/2}\frac{\varepsilon}{t}=0~,
\end{align*}
as required. This completes the proof of (\ref{uas}).

\subsection{Asymptotic behavior of $v$}

Let%
\[
V(k,t)=V_{0}(k,t)+V_{1}(k,t)~,
\]
where%
\begin{align*}
V_{0}(k,t)  & =\left(  \omega_{-}^{\ast}(k)-\frac{1}{\Lambda_{0}}\int_{1}%
^{t}q_{1}(k,s)~ds\right)  (-ik)e^{\Lambda_{-}(t-1)}~,\\
V_{1}(k,t)  & =\left(  i\sigma(k)u_{-}^{\ast}(k)-\frac{1}{2}i\sigma(k)\int
_{1}^{t}q_{1}(k,s)~ds-\frac{1}{2}\int_{1}^{t}q_{0,0}(k,s)~ds\right)
e^{-\left\vert k\right\vert (t-1)}~,
\end{align*}
with $u_{-}^{\ast}$ given by (\ref{um}). Using the triangle inequality we get
that%
\begin{align}
\left\vert v(k,t)-v_{as}(k,t)\right\vert  & \leq\left\vert V_{0}%
(k,t)-c~ike^{-k^{2}t}\right\vert \nonumber\\
& +\left\vert V_{1}(k,t)-\left(  d~i\sigma(k)e^{-\left\vert k\right\vert
t}-b~e^{-\left\vert k\right\vert t}\right)  \right\vert \nonumber\\
& +\left\vert v(k,t)-V(k,t)\right\vert ~.\label{vt}%
\end{align}
We bound each term in (\ref{vt}) separately. First, we have that%
\begin{equation}
\lim_{t\rightarrow\infty}t^{1/2}V_{0}(\frac{k}{t^{1/2}},t)=c~(-ik)e^{-k^{2}%
}~,\label{ldv1}%
\end{equation}
and furthermore that%
\begin{align*}
\left\vert V_{0}(k,t)\right\vert  & \leq\left(  \varepsilon\mu_{\alpha
+1}(k,1)+\varepsilon\mu_{\alpha+1}(k,1)\int_{1}^{\infty}\frac{1}{s^{3/2}%
}ds\right)  \left\vert k\right\vert e^{\Lambda_{-}(t-1)}\\
& \leq\varepsilon\left\vert k\right\vert \mu_{\alpha+1}(k,t)~,
\end{align*}
so that%
\begin{equation}
\left\vert t^{1/2}V_{0}(\frac{k}{t^{1/2}},t)\right\vert \leq\varepsilon
\left\vert k\right\vert \mu_{\alpha+1}(k,1)~.\label{ldv2}%
\end{equation}
From (\ref{ldv1}) and (\ref{ldv2}) it follows by the Lebesgue dominated
convergence theorem that%
\begin{align*}
& \lim_{t\rightarrow\infty}t\int_{\mathbf{R}}\left\vert V_{0}%
(k,t)-c~(-ik)e^{-k^{2}}\right\vert ~dk\\
& =\lim_{t\rightarrow\infty}\int_{\mathbf{R}}\left\vert t^{1/2}V_{0}(\frac
{k}{t^{1/2}},t)-c~(-ik)e^{-k^{2}}\right\vert ~dk\\
& =0~,
\end{align*}
as required. Similarly, since $q_{0}(0,t)=q_{0,0}(0,t)$, for $t\geq1$, we find
that
\begin{equation}
\lim_{t\rightarrow\infty}V_{1}(\frac{k}{t},t)=d~i\sigma(k)e^{-\left\vert
k\right\vert t}-b~e^{-\left\vert k\right\vert t}~,\label{ldv3}%
\end{equation}
and furthermore that%
\begin{align*}
\left\vert V_{1}(k,t)\right\vert  & \leq\left(  \varepsilon\mu_{\alpha
+1}(k,1)+\varepsilon\mu_{\alpha}(k,1)\int_{1}^{\infty}\frac{1}{s^{3/2}}%
ds+\mu_{\alpha+1}(k,1)\left\vert k\right\vert \int_{1}^{t}\frac{1}{s^{1/2}%
}ds\right)  e^{-\left\vert k\right\vert (t-1)}~,\\
& \leq\varepsilon\mu_{\alpha}(k,1)e^{-\left\vert k\right\vert (t-1)}%
+t^{1/2}\mu_{\alpha+1}(k,1)e^{-\left\vert k\right\vert (t-1)}\left\vert
k\right\vert \left(  \frac{t-1}{t}\right) \\
& \leq\varepsilon\bar{\mu}_{\alpha}(k,t)~,
\end{align*}
so that%
\begin{equation}
\left\vert V_{1}(\frac{k}{t},t)\right\vert \leq\varepsilon\mu_{\alpha
}(k,1)~,\label{dv4}%
\end{equation}
for $t\geq1$. From (\ref{ldv3}) and (\ref{dv4}) it follows by the Lebesgue
dominated convergence theorem that%
\begin{align*}
& \lim_{t\rightarrow\infty}t\int_{\mathbf{R}}\left\vert V_{1}(k,t)-\left(
d~i\sigma(k)e^{-\left\vert k\right\vert t}-b~e^{-\left\vert k\right\vert
t}\right)  \right\vert ~dk\\
& =\lim_{t\rightarrow\infty}\int_{\mathbf{R}}\left\vert V_{1}(\frac{k}%
{t},t)-\left(  d~i\sigma(k)e^{-\left\vert k\right\vert t}-b~e^{-\left\vert
k\right\vert t}\right)  \right\vert ~dk\\
& =0~,
\end{align*}
as required. Finally, for the last term in (\ref{vt}) we have the following Proposition:

\begin{proposition}
\label{last}Let $v$ and $V$ be as defined above. Then,%
\begin{align}
\left\vert v(k,t)-V(k,t)\right\vert  & \leq\frac{\varepsilon}{t^{2/3}}%
\mu_{\alpha}(k,t)+\frac{\varepsilon}{t^{1/3}}\mu_{\alpha}^{5/6}%
(k,t)\nonumber\\
& +\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}^{3/4}(k,t)+\frac{\varepsilon
}{t^{1/2}}\bar{\mu}_{\alpha}(k,t)~.\label{vvp}%
\end{align}

\end{proposition}

See Appendix V for a proof.

\bigskip

From Proposition \ref{last} it follows that%
\begin{align*}
& \lim_{t\rightarrow\infty}t\int_{\mathbf{R}}\left\vert
v(k,t)-V(k,t)\right\vert ~dk\\
& \leq\lim_{t\rightarrow\infty}t\int_{\mathbf{R}}\left(  \frac{\varepsilon
}{t^{2/3}}\mu_{\alpha}(k,t)+\frac{\varepsilon}{t^{1/3}}\mu_{\alpha}%
^{5/6}(k,t)+\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}^{3/4}(k,t)+\frac
{\varepsilon}{t^{1/2}}\bar{\mu}_{\alpha+1}(k,t)\right)  ~dk\\
& \leq\lim_{t\rightarrow\infty}t\frac{\varepsilon}{t^{7/6}}=0~,
\end{align*}
as required. This completes the proof of (\ref{vas}).

\section{Appendix V}

In this appendix we prove Proposition \ref{last}. The proof is rather lengthy
and we therefore split it in several pieces. We start by proving some general bound.

\subsection{Two inequalities}

\begin{proposition}
\label{ppp}Let $\alpha\geq0$. Then,%
\begin{align}
\int_{1}^{t}\left(  e^{-\left\vert k\right\vert (t-s)}-e^{-\left\vert
k\right\vert (t-1)}\right)  ~\frac{1}{s^{3/2}}\mu_{\alpha}(k,s)~ds  &
\leq\mathrm{const.}~\left(  \frac{1}{t^{2/3}}\mu_{\alpha}(k,t)+\frac
{1}{t^{1/3}}\mu_{\alpha}^{5/6}(k,t)\right)  ~,\label{ppp1}\\
\int_{t}^{\infty}e^{\left\vert k\right\vert (t-s)}~\frac{1}{s^{3/2}}%
\mu_{\alpha}(k,s)~ds  & \leq\mathrm{const.}~\left(  \frac{1}{t^{2/3}}%
\mu_{\alpha}(k,t)+\frac{1}{t^{1/2}}\mu_{\alpha}^{3/4}(k,t)\right)
~,\label{ppp2}%
\end{align}
uniformly in $k\in\mathbf{R}$ and $t\geq1$.
\end{proposition}

We first prove (\ref{ppp1}) for $1\leq t\leq2$. We have%
\begin{align*}
\int_{1}^{t}\left(  e^{-\left\vert k\right\vert (t-s)}-e^{-\left\vert
k\right\vert (t-1)}\right)  \frac{1}{s^{3/2}}\mu_{\alpha}(k,s)~ds  &
\leq\varepsilon\mu_{\alpha}(k,1)\int_{1}^{t}\frac{ds}{s^{3/2}}\\
& \leq\varepsilon\mu_{\alpha}(k,1)\leq\frac{\varepsilon}{t^{2/3}}\mu_{\alpha
}(k,t)~,
\end{align*}
as required. For $t>2$ we split the integral in (\ref{ppp1}) into two. For the
first part we have%
\begin{align*}
& \int_{1}^{t-(t-1)^{5/6}}\left(  e^{-\left\vert k\right\vert (t-s)}%
-e^{-\left\vert k\right\vert (t-1)}\right)  \frac{1}{s^{3/2}}\mu_{\alpha
}(k,s)~ds\\
& \leq\varepsilon\mu_{\alpha}(k,1)\int_{1}^{t-(t-1)^{5/6}}\left(
e^{-\left\vert k\right\vert (t-s)}-e^{-\left\vert k\right\vert (t-1)}\right)
\frac{1}{s^{3/2}}~ds\\
& \leq\varepsilon\mu_{\alpha}(k,1)\int_{1}^{t-(t-1)^{5/6}}e^{-\left\vert
k\right\vert (t-s)}\left(  1-e^{\left\vert k\right\vert (s-1)}\right)
\frac{1}{s^{3/2}}~ds\\
& \leq\varepsilon\mu_{\alpha}(k,1)e^{-\left\vert k\right\vert (t-1)^{5/6}}%
\int_{1}^{t-(t-1)^{5/6}}(s-1)\left\vert k\right\vert \frac{1}{s^{3/2}}~ds\\
& \leq\varepsilon t^{1/2}\mu_{\alpha}(k,1)e^{-\left\vert k\right\vert
(t-1)^{5/6}}\left\vert k\right\vert \left(  \frac{t-1}{t}\right)  ^{2}\\
& \leq\frac{\varepsilon}{t^{1/3}}\mu_{\alpha}^{5/6}(k,t)~,
\end{align*}
as required, and for the other part we get,%
\begin{align*}
& \int_{t-(t-1)^{5/6}}^{t}\left(  e^{-\left\vert k\right\vert (t-s)}%
-e^{-\left\vert k\right\vert (t-1)}\right)  \frac{1}{s^{3/2}}\mu_{\alpha
}(k,s)~ds\\
& \leq\frac{\varepsilon}{t^{3/2}}\mu_{\alpha}(k,t)\int_{t-(t-1)^{5/6}}^{t}ds\\
& \leq\frac{\varepsilon}{t^{3/2}}\mu_{\alpha}(k,t)t^{5/6}\leq\frac
{\varepsilon}{t^{2/3}}\mu_{\alpha}(k,t)~,
\end{align*}
as required. We now prove (\ref{ppp2}). Namely,%
\begin{align*}
& \int_{t}^{\infty}e^{\left\vert k\right\vert (t-s)}\frac{1}{s^{3/2}}%
\mu_{\alpha}(k,s)~ds\\
& \leq\varepsilon\mu_{\alpha}(k,t)\left(  \int_{t}^{t+t^{3/4}}e^{\left\vert
k\right\vert (t-s)}\frac{1}{s^{3/2}}~ds+\int_{t+t^{3/4}}^{\infty}e^{\left\vert
k\right\vert (t-s)}\frac{1}{s^{3/2}}~ds\right) \\
& \leq\varepsilon\mu_{\alpha}(k,t)\left(  \int_{t}^{t+t^{3/4}}\frac{1}%
{s^{3/2}}~ds+e^{-\left\vert k\right\vert t^{3/4}}\int_{t+t^{3/4}}^{\infty
}\frac{1}{s^{3/2}}~ds\right) \\
& \leq\varepsilon\mu_{\alpha}(k,t)\left(  \frac{1}{t^{3/4}}+\frac{1}{t^{1/2}%
}e^{-\left\vert k\right\vert t^{3/4}}\right) \\
& \leq\frac{\varepsilon}{t^{3/4}}\mu_{\alpha}(k,t)+\frac{\varepsilon}{t^{1/2}%
}\mu_{\alpha}^{3/4}(k,t)\\
& \leq\frac{\varepsilon}{t^{2/3}}\mu_{\alpha}(k,t)+\frac{\varepsilon}{t^{1/2}%
}\mu_{\alpha}^{3/4}(k,t)~,
\end{align*}
as required. This completes the proof of Proposition \ref{ppp}.

\subsection{Proof of Proposition \ref{last}}

Let $v_{D}=v-V$. Using the definitions we find that
\begin{align}
v_{D}(k,t)  & =\omega_{1}(k,t)\nonumber\\
& +\frac{1}{\Lambda_{0}}\int_{1}^{t}\left(  e^{\Lambda_{-}(t-s)}%
-e^{\Lambda_{-}(t-1)}\right)  ikq_{1}(k,s)~ds\nonumber\\
& -\frac{1}{2}\int_{1}^{t}\left(  e^{-\left\vert k\right\vert (t-s)}%
-e^{-\left\vert k\right\vert (t-1)}\right)  \left(  q_{0,0}(k,s)+i\sigma
(k)q_{1}(k,s)\right)  ~ds\nonumber\\
& +\frac{1}{2}\int_{t}^{\infty}e^{\left\vert k\right\vert (t-s)}\left(
q_{0,0}(k,s)-i\sigma(k)q_{1}(k,s)\right)  ~ds\nonumber\\
& -\frac{1}{2}\left(  \int_{1}^{t}e^{-\left\vert k\right\vert (t-s)}%
ikq_{0,1}(k,s)~ds-\int_{t}^{\infty}e^{\left\vert k\right\vert (t-s)}%
ikq_{0,1}(k,s)~ds\right)  ~.\label{vv}%
\end{align}
We write $v_{D}=\sum_{i=1}^{5}v_{D,i}$, with $v_{D,i}$ the $i$-th term in
(\ref{vv}), and we bound each of the terms individually. The inequality
(\ref{vvp}) then follows using the triangle inequality.

\begin{proposition}
\label{v1p}For all $\alpha\geq0$ we have the bounds%
\begin{align}
\left\vert v_{D,1}(k,t)\right\vert  & \leq\frac{\varepsilon}{t}\mu_{\alpha
}(k,t)~,\label{v1b1}\\
\left\vert v_{D,2}(k,t)\right\vert  & \leq\frac{\varepsilon}{t}\mu_{\alpha
}(k,t)~,\label{v1b2}\\
\left\vert v_{D,3}(k,t)\right\vert  & \leq\frac{\varepsilon}{t^{2/3}}%
\mu_{\alpha}(k,t)+\frac{\varepsilon}{t^{1/3}}\mu_{\alpha}^{5/6}%
(k,t)~,\label{v1b3}\\
\left\vert v_{D,4}(k,t)\right\vert  & \leq\frac{\varepsilon}{t^{2/3}}%
\mu_{\alpha}(k,t)+\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}^{3/4}%
(k,t)~,\label{v1b4}\\
\left\vert v_{D,5}(k,t)\right\vert  & \leq\frac{\varepsilon}{t^{1/2}}\bar{\mu
}_{\alpha+1}(k,t)+\frac{\varepsilon}{t^{2/3}}\mu_{\alpha}(k,t)\nonumber\\
& +\frac{\varepsilon}{t^{1/3}}\mu_{\alpha}^{5/6}(k,t)+\frac{\varepsilon
}{t^{1/2}}\mu_{\alpha}^{3/4}(k,t)~,\label{v1b5}%
\end{align}
uniformly in $k\in\mathbf{R}$ and $t\geq1$.
\end{proposition}

The bound on $v_{D,1}$ has already been established in (\ref{bpm2}). Next, to
bound $v_{D,2}$, we split the integral into two parts. We have%
\begin{align*}
& \frac{1}{\Lambda_{0}}\int_{1}^{\frac{t+1}{2}}\left(  e^{\Lambda_{-}%
(t-s)}-e^{\Lambda_{-}(t-1)}\right)  \left\vert k\right\vert \left\vert
q_{1}(k,s)\right\vert ~ds\\
& \leq\varepsilon\mu_{\alpha+1/2}(k,1)\int_{1}^{\frac{t+1}{2}}\left(
e^{\Lambda_{-}(t-s)}-e^{\Lambda_{-}(t-1)}\right)  \left\vert \Lambda
_{-}\right\vert ^{1/2}\frac{1}{s^{3/2}}~ds\\
& \leq\varepsilon\mu_{\alpha+1/2}(k,1)\int_{1}^{\frac{t+1}{2}}e^{\Lambda
_{-}(t-s)}\left(  1-e^{\Lambda_{-}(s-1)}\right)  \left\vert \Lambda
_{-}\right\vert ^{1/2}~\frac{1}{s^{3/2}}~ds\\
& \leq\varepsilon\mu_{\alpha+1/2}(k,1)e^{\Lambda_{-}\frac{t-1}{2}}\int
_{1}^{\frac{t+1}{2}}(s-1)\left\vert \Lambda_{-}\right\vert ^{3/2}\frac
{1}{s^{3/2}}~ds\\
& \leq\varepsilon t^{1/2}\mu_{\alpha+1/2}(k,1)e^{\Lambda_{-}\frac{t-1}{2}%
}\left\vert \Lambda_{-}\right\vert ^{3/2}\left(  \frac{t-1}{t}\right)  ^{2}\\
& \leq\frac{\varepsilon}{t}\mu_{\alpha+1/2}(k,t)~,
\end{align*}
and%
\begin{align*}
& \frac{1}{\Lambda_{0}}\int_{\frac{t+1}{2}}^{t}\left(  e^{\Lambda_{-}%
(t-s)}-e^{\Lambda_{-}(t-1)}\right)  \left\vert k\right\vert \left\vert
q_{1}(k,s)\right\vert ~ds\\
& \leq\varepsilon\frac{\Lambda_{+}^{1/2}}{\Lambda_{0}}\mu_{\alpha}%
(k,t)\int_{\frac{t+1}{2}}^{t}e^{\Lambda_{-}(t-s)}\left\vert \Lambda
_{-}\right\vert ^{1/2}\frac{1}{s^{3/2}}~ds\\
& +\varepsilon\frac{\Lambda_{+}^{1/2}}{\Lambda_{0}}\mu_{\alpha}(k,t)e^{\Lambda
_{-}(t-1)}\left\vert \Lambda_{-}\right\vert ^{1/2}\int_{\frac{t+1}{2}}%
^{t}\frac{1}{s^{3/2}}~ds\\
& \leq\frac{\varepsilon}{t^{3/2}}\mu_{\alpha}(k,t)\int_{\frac{t+1}{2}}%
^{t}e^{\Lambda_{-}(t-s)}\left\vert \Lambda_{-}\right\vert ^{1/2}~ds\\
& +\frac{\varepsilon}{t^{1/2}}\mu_{\alpha}(k,t)e^{\Lambda_{-}(t-1)}\left\vert
\Lambda_{-}\right\vert ^{1/2}\left(  \frac{t-1}{t}\right) \\
& \leq\frac{\varepsilon}{t^{3/2}}\mu_{\alpha}(k,t)\int_{\frac{t+1}{2}}%
^{t}\frac{1}{\sqrt{t-s}}~ds+\frac{\varepsilon}{t}\mu_{\alpha}(k,t)\\
& \leq\frac{\varepsilon}{t}\mu_{\alpha}(k,t)~,
\end{align*}
and (\ref{v1b2}) now follows using the triangle inequality. The bounds
(\ref{v1b3}) and (\ref{v1b4}) on $v_{D,3}$ and $v_{D,4}$ are a consequence of
Proposition \ref{ppp}, using that%
\[
\left\vert q_{0,0}(k,s)+i\sigma(k)q_{1}(k,s)\right\vert \leq\frac{\varepsilon
}{s^{3/2}}\mu_{\alpha}(k,s)~.
\]
To complete the proof of Proposition \ref{v1p} we still need to prove the
bound (\ref{v1b5}) on $v_{D,5}$. This bound is somewhat tricky, since the
dominant contributions to the two integrals defining $v_{D,5}$ compensate one
another. A proof of this bound is the content of the final subsections of this paper.

\subsection{Proof of inequality (\ref{v1b5})}

Using the definition of $v_{D,5}$ we find after integration by parts, that%
\begin{align*}
-i\sigma(k)v_{D,5}(k,t)  & =\int_{1}^{t}e^{-\left\vert k\right\vert
(t-s)}\left\vert k\right\vert q_{0,1}(k,s)~ds-\int_{t}^{\infty}e^{\left\vert
k\right\vert (t-s)}\left\vert k\right\vert q_{0,1}(k,s)~ds\\
& =\left[  e^{-\left\vert k\right\vert (t-s)}q_{0,1}(k,s)\right]  _{s=1}%
^{s=t}-\int_{1}^{t}e^{-\left\vert k\right\vert (t-s)}\partial_{s}%
q_{0,1}(k,s)~ds\\
& +\left[  e^{\left\vert k\right\vert (t-s)}q_{0,1}(k,s)\right]
_{s=t}^{s=\infty}-\int_{t}^{\infty}e^{\left\vert k\right\vert (t-s)}%
\partial_{s}q_{0,1}(k,s)~ds~.
\end{align*}
Therefore,%
\begin{align*}
-i\sigma(k)v_{D,5}(k,t)  & =q_{0,1}(k,t)-e^{-\left\vert k\right\vert
(t-1)}q_{0,1}(k,1)-q_{0,1}(k,t)\\
& -e^{-\left\vert k\right\vert (t-1)}\int_{1}^{t}\partial_{s}q_{0,1}(k,s)~ds\\
& -\int_{1}^{t}\left(  e^{-\left\vert k\right\vert (t-s)}-e^{-\left\vert
k\right\vert (t-1)}\right)  \partial_{s}q_{0,1}(k,s)~ds\\
& -\int_{t}^{\infty}e^{\left\vert k\right\vert (t-s)}\partial_{s}%
q_{0,1}(k,s)~ds~,
\end{align*}
and therefore%
\begin{align}
v_{D,5}(k,t)  & =-i\sigma(k)e^{-\left\vert k\right\vert (t-1)}q_{0,1}%
(k,t)\nonumber\\
& -i\sigma(k)\int_{1}^{t}\left(  e^{-\left\vert k\right\vert (t-s)}%
-e^{-\left\vert k\right\vert (t-1)}\right)  \partial_{s}q_{0,1}%
(k,s)~ds\nonumber\\
& -i\sigma(k)\int_{t}^{\infty}e^{\left\vert k\right\vert (t-s)}\partial
_{s}q_{0,1}(k,s)~ds~.\label{v15}%
\end{align}
From the representation (\ref{v15}) of $v_{D,5}$ we get the inequality%
\begin{align}
\left\vert v_{D,5}(k,t)\right\vert  & \leq e^{-\left\vert k\right\vert
(t-1)}\left\vert q_{0,1}(k,t)\right\vert \nonumber\\
& +\int_{1}^{t}\left(  e^{-\left\vert k\right\vert (t-s)}-e^{-\left\vert
k\right\vert (t-1)}\right)  \left\vert \partial_{s}q_{0,1}(k,s)\right\vert
~ds\nonumber\\
& +\int_{t}^{\infty}e^{\left\vert k\right\vert (t-s)}\left\vert \partial
_{s}q_{0,1}(k,s)\right\vert ~ds~.\label{v1b}%
\end{align}
We now prove a bound for each of the terms in (\ref{v1b}) separately. The
bound (\ref{v1b5}) on $v_{D,5}$ then follows by the triangle inequality. For
the first term in (\ref{v1b}) we have that%
\begin{align*}
e^{-\left\vert k\right\vert (t-1)}\left\vert q_{0,1}(k,t)\right\vert  & \leq
e^{-\left\vert k\right\vert (t-1)}\frac{\varepsilon}{t^{1/2}}\mu_{\alpha
+1}(k,t)\\
& \leq\frac{\varepsilon}{t^{1/2}}\bar{\mu}_{\alpha+1}(k,t)~,
\end{align*}
as required, and the bounds on the second and the third term in (\ref{v1b})
follow using Proposition \ref{ppp}, together with:

\begin{proposition}
\label{pppp0}For all $\alpha\geq0$ we have that%
\begin{equation}
\left\vert \partial_{t}q_{0,1}(k,t)\right\vert \leq\frac{\varepsilon}{t^{3/2}%
}\mu_{\alpha}(k,t)~,\label{dtq0b}%
\end{equation}
uniformly in $k\in\mathbf{R}$ and $t\geq1$.
\end{proposition}

See the next subsection for a proof.

\bigskip

This completes the proof of inequality (\ref{v1b5}).

\subsection{Proof of Proposition \ref{pppp0}}

By definition (\ref{q01}) of $q_{0,1}$ we have that%
\[
\partial_{t}q_{0,1}(k,t)=-\frac{1}{2\pi}\left(  u_{0}\ast\partial_{t}%
u_{0}\right)  (k,t)~,
\]
and for $\partial_{t}u_{0}$ we have:

\begin{proposition}
\label{pppp1}For all $\alpha\geq0$ we have%
\begin{equation}
\left\vert \partial_{t}u_{0}(k,t)\right\vert \leq\frac{\varepsilon}{t}%
\mu_{\alpha}(k,t)~,\label{dtu0bb}%
\end{equation}
uniformly in $k\in\mathbf{R}$ and $t\geq1$.
\end{proposition}

See the next subsection for a proof.

\bigskip

Using (\ref{c1b4}) we find from (\ref{bpm3}) and (\ref{dtu0bb}) that
\[
\left\vert \left(  u_{0}\ast\partial_{t}u_{0}\right)  (k,t)\right\vert
\leq\frac{\varepsilon^{2}}{t^{3/2}}\mu_{a}(k,t)~,
\]
and (\ref{dtq0b}) follows. This completes the proof of Proposition \ref{pppp0}.

\subsection{Proof of Proposition \ref{pppp1}}

By definition (\ref{u0}) of $u_{0}$ we have that
\begin{align}
\partial_{t}u_{0}(k,t)  & =-\omega_{-}^{\ast}(k)\Lambda_{-}e^{\Lambda
_{-}(t-1)}\nonumber\\
& +\frac{1}{\Lambda_{0}}q_{1}(k,t)\nonumber\\
& +\frac{1}{\Lambda_{0}}\int_{1}^{t}e^{\Lambda_{-}(t-s)}\Lambda_{-}%
q_{1}(k,s)~ds~.\label{dtu0}%
\end{align}
We write $\partial_{t}u_{0}=\sum_{i=1}^{3}\partial_{t}u_{0,i}$, with
$\partial_{t}u_{0,i}$ the $i$-th term in (\ref{dtu0}), and we bound each of
the terms separately. The inequality (\ref{dtu0bb}) then follows using the
triangle inequality.

\begin{proposition}
\label{pppp}For all $\alpha\geq0$ we have the bounds%
\begin{align}
\left\vert \partial_{t}u_{0,1}(k,t)\right\vert  & \leq\frac{\varepsilon}{t}%
\mu_{a}(k,t)~,\label{dtu0b1}\\
\left\vert \partial_{t}u_{0,2}(k,t)\right\vert  & \leq\frac{\varepsilon
}{t^{3/2}}\mu_{a}(k,t)~,\label{dtu0b2}\\
\left\vert \partial_{t}u_{0,3}(k,t)\right\vert  & \leq\frac{\varepsilon}{t}%
\mu_{a+1}(k,t)~,\label{dtu0b3}%
\end{align}
uniformly in $t\geq1$ and $k\in\mathbf{R}$.
\end{proposition}

For $\partial_{t}u_{0,1}$ we have%
\[
\left\vert -\omega_{-}^{\ast}(k)\Lambda_{-}e^{\Lambda_{-}(t-1)}\right\vert
\leq\varepsilon\mu_{a+1}(k,1)e^{\Lambda_{-}(t-1)}\left\vert \Lambda
_{-}\right\vert ~,
\]
and (\ref{dtu0b1}) follows using Proposition \ref{basic}. Next,
\[
\left\vert \frac{1}{\Lambda_{0}}q_{1}(k,t)\right\vert \leq\frac{1}{\Lambda
_{0}}\frac{\varepsilon}{t^{3/2}}\mu_{a}(k,t)~,
\]
and (\ref{dtu0b2}) follows. Finally, splitting the integral defining
$\partial_{t}u_{0,3}$ in two parts we find that%
\begin{align*}
\left\vert \frac{1}{\Lambda_{0}}\int_{1}^{\frac{t+1}{2}}e^{\Lambda_{-}%
(t-s)}\Lambda_{-}q_{1}(k,s)~ds\right\vert  & \leq\varepsilon\mu_{a+1}%
(k,1)e^{\Lambda_{-}\frac{t-1}{2}}\left\vert \Lambda_{-}\right\vert \int
_{1}^{\frac{t+1}{2}}\frac{1}{s^{3/2}}ds\\
& \leq\varepsilon\mu_{a+1}(k,1)e^{\Lambda_{-}\frac{t-1}{2}}\left\vert
\Lambda_{-}\right\vert \left(  \frac{t-1}{t}\right) \\
& \leq\frac{\varepsilon}{t}\mu_{a+1}(k,t)~,
\end{align*}
and that%
\begin{align*}
\left\vert \frac{1}{\Lambda_{0}}\int_{\frac{t+1}{2}}^{t}e^{\Lambda_{-}%
(t-s)}\left\vert \Lambda_{-}\right\vert q_{1}(k,s)~ds\right\vert  & \leq
\frac{\varepsilon}{t^{3/2}}\frac{1}{\Lambda_{0}}\mu_{a}(k,t)\int_{\frac
{t+1}{2}}^{t}e^{\Lambda_{-}(t-s)}\left\vert \Lambda_{-}\right\vert ds\\
& \leq\frac{\varepsilon}{t^{3/2}}\frac{1}{\Lambda_{0}}\mu_{a}(k,t)\\
& \leq\frac{\varepsilon}{t}\mu_{a+1}(k,t)~,
\end{align*}
and (\ref{dtu0b3}) follows using the triangle inequality. This completes the
proof of Proposition \ref{pppp} and Proposition \ref{pppp1}.

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